16x^2+48x-256=0

Solution for 16x^2+48x-256=0 equation:

16x^2+48x-256=0

a = 16; b = 48; c = -256;
Δ = b2-4ac
Δ = 482-4·16·(-256)
Δ = 18688
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:

$\sqrt{\Delta}=\sqrt{18688}=\sqrt{256*73}=\sqrt{256}*\sqrt{73}=16\sqrt{73}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-16\sqrt{73}}{2*16}=\frac{-48-16\sqrt{73}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+16\sqrt{73}}{2*16}=\frac{-48+16\sqrt{73}}{32}
$