4x^2+40x=-90

Solution for 4x^2+40x=-90 equation:

4x^2+40x=-90

We move all terms to the left:

4x^2+40x-(-90)=0

We add all the numbers together, and all the variables

4x^2+40x+90=0

a = 4; b = 40; c = +90;
Δ = b2-4ac
Δ = 402-4·4·90
Δ = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-4\sqrt{10}}{2*4}=\frac{-40-4\sqrt{10}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+4\sqrt{10}}{2*4}=\frac{-40+4\sqrt{10}}{8}
$