x^2+48x-512=0

Solution for x^2+48x-512=0 equation:

x^2+48x-512=0

a = 1; b = 48; c = -512;
Δ = b2-4ac
Δ = 482-4·1·(-512)
Δ = 4352
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:

$\sqrt{\Delta}=\sqrt{4352}=\sqrt{256*17}=\sqrt{256}*\sqrt{17}=16\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-16\sqrt{17}}{2*1}=\frac{-48-16\sqrt{17}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+16\sqrt{17}}{2*1}=\frac{-48+16\sqrt{17}}{2}
$