0=-16x^2+40x+72

Solution for 0=-16x^2+40x+72 equation:

0=-16x^2+40x+72

We move all terms to the left:

0-(-16x^2+40x+72)=0

We add all the numbers together, and all the variables

-(-16x^2+40x+72)=0

We get rid of parentheses

16x^2-40x-72=0

a = 16; b = -40; c = -72;
Δ = b2-4ac
Δ = -402-4·16·(-72)
Δ = 6208
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:

$\sqrt{\Delta}=\sqrt{6208}=\sqrt{64*97}=\sqrt{64}*\sqrt{97}=8\sqrt{97}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-8\sqrt{97}}{2*16}=\frac{40-8\sqrt{97}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+8\sqrt{97}}{2*16}=\frac{40+8\sqrt{97}}{32}
$