# How do you find the discriminant for 0.5x^2-2x=-2 and determine the number and type of solutions?

Aug 1, 2017

$D = 0$, so there is one root and the root is $2$

#### Explanation:

$0.5 {x}^{2} - 2 x = - 2 \mathmr{and} 0.5 {x}^{2} - 2 x + 2 = 0$

Comparing with standard quadratic equation $a {x}^{2} + b x + c = 0$

We get here $a = 0.5 , b = - 2 , c = 2$

Discriminant $D = {b}^{2} - 4 a c = 4 - 4 \cdot 0.5 \cdot 2 = 4 - 4 = 0$ We

know if $D > 0$ two real roots , if $D = 0$ one real root. and

if $D < 0$ two complex roots. here $D = 0$ so there is one root.

Root : $x = \frac{- b \pm \sqrt{D}}{2 a} = \frac{2}{2 \cdot 0.5} = \frac{2}{1} = 2$ [Ans]