36x^2+320x+672=0

Solution for 36x^2+320x+672=0 equation:

36x^2+320x+672=0

a = 36; b = 320; c = +672;
Δ = b2-4ac
Δ = 3202-4·36·672
Δ = 5632
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:

$\sqrt{\Delta}=\sqrt{5632}=\sqrt{256*22}=\sqrt{256}*\sqrt{22}=16\sqrt{22}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(320)-16\sqrt{22}}{2*36}=\frac{-320-16\sqrt{22}}{72}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(320)+16\sqrt{22}}{2*36}=\frac{-320+16\sqrt{22}}{72}
$