4x^2+72x=-320

Solution for 4x^2+72x=-320 equation:

4x^2+72x=-320

We move all terms to the left:

4x^2+72x-(-320)=0

We add all the numbers together, and all the variables

4x^2+72x+320=0

a = 4; b = 72; c = +320;
Δ = b2-4ac
Δ = 722-4·4·320
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(72)-8}{2*4}=\frac{-80}{8}
=-10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(72)+8}{2*4}=\frac{-64}{8}
=-8
$