**Calculate second degree equations online** is possible thanks to our calculator, just enter the value of the coefficients a, b and c to obtain the possible solutions to the equation. Depending on the value of the discriminant **(b ^{2}-4ac), **the roots of the second degree equation can be very different:

**Negative discriminant**The equation has no real solutions.**Discriminant equal to zero**The equation has a double solution**Discriminant greater than zero**The equation has two real solutions with different numbers.

Article sections

## Formula for quadratic equation

Now that we know that a quadratic equation looks similar to the formula above, it is time to learn how to solve it to get the x values.

To do so, we only have to raise and r**esolve the following formula** to obtain the solutions to our quadratic function:

Remember that we will have as a result two solutions for the unknown, hence before the square root we have a + and - symbol.

## Incomplete quadratic equations

If you have an e**Incomplete quadratic equation** then you will not be able to apply the above formula because you will not know the value of any of the coefficients. Depending on each case, you have to proceed in one way or another, so we explain it below:

**ax**^{2} = 0

^{2}= 0

In this case, **x = 0** since for any non-zero value of a, it is the only number that will satisfy this equality. Example:

3x^{2} = 0 → x = 0

**ax**^{2} + bx = 0

^{2}+ bx = 0

This exercise of incomplete second degree equation is very common and is solved in a very simple way. We simply have to take out the common factor and we will have that:

x (ax + b) = 0

From here we can separate the formula into two different equations that will give us the solutions of x:

- x = 0
- ax + b = 0

For example, imagine that we have x^{2} - 5x = 0. Si sacamos factor común tenemos que:

x(x - 5) = 0

And finally:

- x = 0
- x - 5 = 0 → x = 5

**ax**^{2} + c = 0

^{2}+ c = 0

The third and last case of **incomplete quadratic equation** is easy to do. We simply have to clear the value of x, so we are going to pass as many terms as we can to the right to leave the unknown alone. This can be summarized in the following steps:

- ax
^{2}= -c - x
^{2}= -c/a

Finally **we clear the quadratic term** and we will have that the two solutions for x are obtained by performing the following two operations:

This is likely to be of great help to you in resolving biquadratic equations.

## How the quadratic equation calculator works

Our **quadratic equations calculator online** has been designed so that you can solve them immediately, either because you want to save yourself the calculations or because you want to check that you have done the operations correctly.

Its operation is very simple as you have been able to see. As you have been able to see throughout the post, an equation of degree two looks like this:

ax

^{2}+ bx + c = 0

Remember that for the equation to be of second degree, **el valor de 'a' debe ser distinto de cero**.

Therefore, the only thing you need to **solving quadratic equations** with our calculator is to enter the values of a, b and c. Then press the calculate button and you will get the result of the quadratic equation instantly.

## Solving quadratic equations in Excel

If you want to use Excel to **solving quadratic equations** you will have to follow the instructions that we tell you now.

Lo primero de todo es abrir el programa de hojas de cálculo de Microsoft, crear una nueva hoja de cálculo y escribir en las celdas A2, B2 y C2 el valor de los coeficientes 'a', 'b' 'c' de la **quadratic equation** you want to solve.

Now we are going to write the formula that will calculate the possible **solutions of x satisfying the quadratic equation** that we have placed in Excel. To do this, we write this formula in cell B5:

=(-B2+SQRT(B2*B2-4*A2*C2))/(2*A2)

And this other formula in cell B6:

=(-B2-SQRT(B2*B2-4*A2*C2))/(2*A2)

When you have done so, the **Excel to solve second degree equations** will look more or less like the image we have attached a few lines above.

Now you only have to change the values of cells A2, B2 and C2 to obtain the** solutions of all second degree equations** that can be solved using the general formula we have seen in the first point of the theory.

## Examples of quadratic equations

Below we have compiled some of them **examples of second degree equations to print** and solve them at your leisure. Then you can check with our calculator that the solutions you have obtained are correct and find out if you have made a mistake.

- x
^{2}+ 4x + -21 = 0 - 9x
^{2}+ -12x + 4 = 0 - 4x
^{2}+ 28x + 49 = 0 - 4x
^{2}+ -20x + 25 = 0 - x
^{2}+ 9x + 20 = 0 - x
^{2}+ x + 3 = 0 - x
^{2}+ -2x + 3 = 0 - -2x
^{2}+ 3x + 2 = 0

## Quadratic equations solved step by step

In case you still have any doubts, here are the first three **second degree equations solved step by step**:

If you are still having problems with **solving quadratic equations**, or if you have any questions, leave us a comment and we will try to help you as soon as possible.

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Very interesting how it solves equations.

I love it, I can now calculate equations of degree two without applying the formula.

I don't have to solve the second degree formula anymore hahahahahahaha

Guys help me with this one please x 2+6x-7=0

Hi Mariela, the solutions of the quadratic equation you have put are 1 and -7.

can you help me to solve this

x²+5x-24=0

Hi Maria,

The solutions of the quadratic equation you have given us are x = 3 and x = -8.

I hope it has been helpful.

x²+4x-8=0

Hello Milagros,

The second degree equation you pose to us has no real solutions since its discriminant is negative. Look:

(b

^{2}-4ac) = 16 - (-32) = -48You would have to use imaginary numbers to solve it.

Greetings!

how do i solve this = (x-2)(x+5)=9x+10

Hello Gustavo,

Just do the operation on the left-hand terms and solve as a normal second degree equation.

That is to say:

(x-2)(x+5)=9x+10

x

^{2}+ 5x -2x -10 = 9x + 10Now we group the terms on the left and we can calculate it:

x

^{2}-6x - 20 = 0Finally, the solutions of this equation of degree two are x1 = 8.38 and x2 = -2.38.

Greetings!

x²-6x+2 but I can't understand why it gives this result.

What is the result?