-(4x)^2+12x+240=0

Solution for -(4x)^2+12x+240=0 equation:

-(4x)^2+12x+240=0

a = -4; b = 12; c = +240;
Δ = b2-4ac
Δ = 122-4·(-4)·240
Δ = 3984
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:

$\sqrt{\Delta}=\sqrt{3984}=\sqrt{16*249}=\sqrt{16}*\sqrt{249}=4\sqrt{249}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{249}}{2*-4}=\frac{-12-4\sqrt{249}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{249}}{2*-4}=\frac{-12+4\sqrt{249}}{-8}
$