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

1 Answer
Feb 20, 2017

Answer:

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