4x^2+40x=-56

Solution for 4x^2+40x=-56 equation:

4x^2+40x=-56

We move all terms to the left:

4x^2+40x-(-56)=0

We add all the numbers together, and all the variables

4x^2+40x+56=0

a = 4; b = 40; c = +56;
Δ = b2-4ac
Δ = 402-4·4·56
Δ = 704
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{704}=\sqrt{64*11}=\sqrt{64}*\sqrt{11}=8\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-8\sqrt{11}}{2*4}=\frac{-40-8\sqrt{11}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+8\sqrt{11}}{2*4}=\frac{-40+8\sqrt{11}}{8}
$