4x^2+40x=-64

Solution for 4x^2+40x=-64 equation:

4x^2+40x=-64

We move all terms to the left:

4x^2+40x-(-64)=0

We add all the numbers together, and all the variables

4x^2+40x+64=0

a = 4; b = 40; c = +64;
Δ = b2-4ac
Δ = 402-4·4·64
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-24}{2*4}=\frac{-64}{8}
=-8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+24}{2*4}=\frac{-16}{8}
=-2
$