-4t^2+16t+15=0

Solution for -4t^2+16t+15=0 equation:

-4t^2+16t+15=0

a = -4; b = 16; c = +15;
Δ = b2-4ac
Δ = 162-4·(-4)·15
Δ = 496
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:

$\sqrt{\Delta}=\sqrt{496}=\sqrt{16*31}=\sqrt{16}*\sqrt{31}=4\sqrt{31}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{31}}{2*-4}=\frac{-16-4\sqrt{31}}{-8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{31}}{2*-4}=\frac{-16+4\sqrt{31}}{-8}
$