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=0

a = 4; b = 72; c = 0;
Δ = b2-4ac
Δ = 722-4·4·0
Δ = 5184
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5184}=72$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(72)-72}{2*4}=\frac{-144}{8}
=-18
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(72)+72}{2*4}=\frac{0}{8}
=0
$