# How do you solve the quadratic 5x^2-3x=-8 using any method?

Feb 20, 2017

no real solutions

#### Explanation:

First, set it to equal to zero.
$5 {x}^{2} - 3 x = - 8$
$5 {x}^{2} - 3 x + 8 = 0$

If factoring the quadratic equation doesn't solve the problem, use the quadratic formula as a last resort to find x$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$ when $a = 5 , b = - 3 , c = 8$

$x = \frac{- \left(- 3\right) \pm \sqrt{{\left(- 3\right)}^{2} - 4 \left(5\right) \left(8\right)}}{2 \left(5\right)} = \frac{3 \pm \sqrt{9 - 160}}{10}$
Now if you look at the discriminant which is ${b}^{2} - 4 a c$, is negative. The discriminant tells you how many real solutions are in the equation.

remember this:
when ${b}^{2} - 4 a c < 0$ is negative , there are no real solutions
when ${b}^{2} - 4 a c = 0$ is zero, there is one real solution
when ${b}^{2} - 4 a c > 0$ is positive, there are two real solutions

so the answer is that there are no real solutions