# How do you find the discriminant for 2x^2-5x+20=0 and determine the number and type of solutions?

Apr 18, 2017

Complex and conjugate. $x = 1.25 + 2.9047 i , x = 1.25 - 2.9047 i$

#### Explanation:

$2 {x}^{2} - 5 x + 20 = 0$. Comparing with standard equation $a {x}^{2} + b x + c = 0$ we get $a = 2 , b = - 5 , c = 20$. Disciminant $D = {b}^{2} - 4 a c = {\left(- 5\right)}^{2} - 4 \cdot 2 \cdot 20 = - 135$ If $D = 0$ roots are equal.
If $D > 0$ The roots are real.
If $D < 0$The roots are complex in nature and conjugate.
Here the discriminant is $< 0$.So roots are complex in nature and conjugate.
$x = - \frac{b}{2 a} \pm \frac{\sqrt{{b}^{2} - 4 a c}}{2 a} = - \frac{- 5}{2 \cdot 2} \pm \frac{\sqrt{{\left(- 5\right)}^{2} - 4 \cdot 2 \cdot 20}}{2 \cdot 2} = 1.25 \pm 2.9047 i$
Solution: $x = 1.25 + 2.9047 i , x = 1.25 - 2.9047 i$ [Ans]