-12x^2+12x+48=0

Solution for -12x^2+12x+48=0 equation:

-12x^2+12x+48=0

a = -12; b = 12; c = +48;
Δ = b2-4ac
Δ = 122-4·(-12)·48
Δ = 2448
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:

$\sqrt{\Delta}=\sqrt{2448}=\sqrt{144*17}=\sqrt{144}*\sqrt{17}=12\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12\sqrt{17}}{2*-12}=\frac{-12-12\sqrt{17}}{-24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12\sqrt{17}}{2*-12}=\frac{-12+12\sqrt{17}}{-24}
$