-16x^2+48x+40=0

Solution for -16x^2+48x+40=0 equation:

-16x^2+48x+40=0

a = -16; b = 48; c = +40;
Δ = b2-4ac
Δ = 482-4·(-16)·40
Δ = 4864
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:

$\sqrt{\Delta}=\sqrt{4864}=\sqrt{256*19}=\sqrt{256}*\sqrt{19}=16\sqrt{19}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-16\sqrt{19}}{2*-16}=\frac{-48-16\sqrt{19}}{-32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+16\sqrt{19}}{2*-16}=\frac{-48+16\sqrt{19}}{-32}
$