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

1 Answer
Feb 20, 2017

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 xx=(-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