4x^2+16x-72=0

Solution for 4x^2+16x-72=0 equation:

4x^2+16x-72=0

a = 4; b = 16; c = -72;
Δ = b2-4ac
Δ = 162-4·4·(-72)
Δ = 1408
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:

$\sqrt{\Delta}=\sqrt{1408}=\sqrt{64*22}=\sqrt{64}*\sqrt{22}=8\sqrt{22}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8\sqrt{22}}{2*4}=\frac{-16-8\sqrt{22}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8\sqrt{22}}{2*4}=\frac{-16+8\sqrt{22}}{8}
$