2x^2+8x-72=0

Solution for 2x^2+8x-72=0 equation:

2x^2+8x-72=0

a = 2; b = 8; c = -72;
Δ = b2-4ac
Δ = 82-4·2·(-72)
Δ = 640
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:

$\sqrt{\Delta}=\sqrt{640}=\sqrt{64*10}=\sqrt{64}*\sqrt{10}=8\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{10}}{2*2}=\frac{-8-8\sqrt{10}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{10}}{2*2}=\frac{-8+8\sqrt{10}}{4}
$