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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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}
$