4x^2-72x+320=96

Solution for 4x^2-72x+320=96 equation:

4x^2-72x+320=96

We move all terms to the left:

4x^2-72x+320-(96)=0

We add all the numbers together, and all the variables

4x^2-72x+224=0

a = 4; b = -72; c = +224;
Δ = b2-4ac
Δ = -722-4·4·224
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-72)-40}{2*4}=\frac{32}{8}
=4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-72)+40}{2*4}=\frac{112}{8}
=14
$