Question

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

Answer 1

no real solutions

Explanation:

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

If factoring the quadratic equation doesn't solve the problem, use the quadratic formula as a last resort to find x`x=(-b +- sqrt(b^2-4ac))/(2a)` when `a=5, b=-3, c=8`

`x=(-(-3) +- sqrt((-3)^2-4(5)(8)))/(2(5))=(3+-sqrt(9-160))/10`
Now if you look at the discriminant which is `b^2-4ac`, is negative. The discriminant tells you how many real solutions are in the equation.

remember this:
when `b^2-4ac<0` is negative , there are no real solutions
when `b^2-4ac=0` is zero, there is one real solution
when `b^2-4ac>0` is positive, there are two real solutions

so the answer is that there are no real solutions