48x^2+48x-95=0

Solution for 48x^2+48x-95=0 equation:

48x^2+48x-95=0

a = 48; b = 48; c = -95;
Δ = b2-4ac
Δ = 482-4·48·(-95)
Δ = 20544
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:

$\sqrt{\Delta}=\sqrt{20544}=\sqrt{64*321}=\sqrt{64}*\sqrt{321}=8\sqrt{321}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-8\sqrt{321}}{2*48}=\frac{-48-8\sqrt{321}}{96}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+8\sqrt{321}}{2*48}=\frac{-48+8\sqrt{321}}{96}
$