3x^2+72x+320=0

Solution for 3x^2+72x+320=0 equation:

3x^2+72x+320=0

a = 3; b = 72; c = +320;
Δ = b2-4ac
Δ = 722-4·3·320
Δ = 1344
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:

$\sqrt{\Delta}=\sqrt{1344}=\sqrt{64*21}=\sqrt{64}*\sqrt{21}=8\sqrt{21}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(72)-8\sqrt{21}}{2*3}=\frac{-72-8\sqrt{21}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(72)+8\sqrt{21}}{2*3}=\frac{-72+8\sqrt{21}}{6}
$