9x^2+48x-64=0

Solution for 9x^2+48x-64=0 equation:

9x^2+48x-64=0

a = 9; b = 48; c = -64;
Δ = b2-4ac
Δ = 482-4·9·(-64)
Δ = 4608
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:

$\sqrt{\Delta}=\sqrt{4608}=\sqrt{2304*2}=\sqrt{2304}*\sqrt{2}=48\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-48\sqrt{2}}{2*9}=\frac{-48-48\sqrt{2}}{18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+48\sqrt{2}}{2*9}=\frac{-48+48\sqrt{2}}{18}
$