-8t^2+36t+15=0

Solution for -8t^2+36t+15=0 equation:

-8t^2+36t+15=0

a = -8; b = 36; c = +15;
Δ = b2-4ac
Δ = 362-4·(-8)·15
Δ = 1776
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1776}=\sqrt{16*111}=\sqrt{16}*\sqrt{111}=4\sqrt{111}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-4\sqrt{111}}{2*-8}=\frac{-36-4\sqrt{111}}{-16}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+4\sqrt{111}}{2*-8}=\frac{-36+4\sqrt{111}}{-16}
$