4x^2+72x+320=3

Solution for 4x^2+72x+320=3 equation:

4x^2+72x+320=3

We move all terms to the left:

4x^2+72x+320-(3)=0

We add all the numbers together, and all the variables

4x^2+72x+317=0

a = 4; b = 72; c = +317;
Δ = b2-4ac
Δ = 722-4·4·317
Δ = 112
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:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(72)-4\sqrt{7}}{2*4}=\frac{-72-4\sqrt{7}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(72)+4\sqrt{7}}{2*4}=\frac{-72+4\sqrt{7}}{8}
$