# Beta version

BETA TEST VERSION OF THIS ITEM

This online calculator is currently under heavy development. It may or it may NOT work correctly.

You CAN try to use it. You CAN even get the proper results.

However, please VERIFY all results on your own, as the level of completion of this item is NOT CONFIRMED.

Feel free to send any ideas and comments !

This online calculator is currently under heavy development. It may or it may NOT work correctly.

You CAN try to use it. You CAN even get the proper results.

However, please VERIFY all results on your own, as the level of completion of this item is NOT CONFIRMED.

Feel free to send any ideas and comments !

# Input equation, which you want to solve

Parameters of the ax^{2} + bx + c = 0 equation | ||

Coefficient a (just before x ^{2}) | ||

Coefficient b (just before x) | ||

Free parameter c |

# The solution of your equation

The equation you entered | ||

Show source$2\cdot{ x}^{2}+5\cdot x - 8 = 0$ | ||

The solution of the equation | ||

Show source$x \in \left\{\frac{-5}{4}-\frac{\sqrt{89}}{4}, \frac{-5}{4}+\frac{\sqrt{89}}{4}\right\}$ |

# The solution step-by-step

I. We calculatate discriminant of the quadratic equation $\Delta$.

$\begin{aligned}\Delta& = {5}^{2} - \left(4\cdot2\cdot\left(-8\right)\right) = 25 - \left(4\cdot2\cdot\left(-8\right)\right) = 25 - \left(-64\right) = \\ & = 25+64 = 89\end{aligned}$

II. Delta is positive (Δ > 0), so equation has two solutions (roots).

The first solution is:

$\begin{aligned}x_1& = \frac{-5 - \sqrt{89}}{2\cdot2} = \frac{-5 - \sqrt{89}}{4} = \frac{-5}{4}-\frac{\sqrt{89}}{4}\end{aligned}$The second solution is:

$\begin{aligned}x_2& = \frac{-5+\sqrt{89}}{2\cdot2} = \frac{-5+\sqrt{89}}{4} = \frac{-5}{4}+\frac{\sqrt{89}}{4}\end{aligned}$

$\begin{aligned}\Delta& = {5}^{2} - \left(4\cdot2\cdot\left(-8\right)\right) = 25 - \left(4\cdot2\cdot\left(-8\right)\right) = 25 - \left(-64\right) = \\ & = 25+64 = 89\end{aligned}$

II. Delta is positive (Δ > 0), so equation has two solutions (roots).

The first solution is:

$\begin{aligned}x_1& = \frac{-5 - \sqrt{89}}{2\cdot2} = \frac{-5 - \sqrt{89}}{4} = \frac{-5}{4}-\frac{\sqrt{89}}{4}\end{aligned}$The second solution is:

$\begin{aligned}x_2& = \frac{-5+\sqrt{89}}{2\cdot2} = \frac{-5+\sqrt{89}}{4} = \frac{-5}{4}+\frac{\sqrt{89}}{4}\end{aligned}$

# Some facts

**The quadratic equation**is an equation that can be presented in the form:

$a~x^2 + b~x + c = 0$where:

**a**,**b**,**c**- constant parameters, these are numbers that we**know**,

**x**-**unknown**variable, it's a number, which we**search**for.

- Quadratic equation can have
**one solution**,**two solutions**or**do not have solutions**. - The universal method of solving quadratic equations uses discriminant of the quadratic polynomial (so-called delta):

$\Delta={ b}^{2}-4~ a~ c$ - When we calculate the discriminant, three scenarios are possible:

**discriminant is positive**(Δ > 0) - equation has**two different solutions**(two different roots):

$x_1=\frac{- b-\sqrt{ \Delta}}{2~ a}$$x_2=\frac{- b+\sqrt{ \Delta}}{2~ a}$**discriminant is zero**(Δ = 0) - equation has exactly**one solution**(so-called double root):

$h=\frac{- b}{2~ a}$**discriminant is negative**(Δ < 0) - equation has**no solutions**(so-called contradictory equation).

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