15t=-16t2+15

Solution for 15t=-16t2+15 equation:

15t=-16t^2+15

We move all terms to the left:

15t-(-16t^2+15)=0

We get rid of parentheses

16t^2+15t-15=0

a = 16; b = 15; c = -15;
Δ = b2-4ac
Δ = 152-4·16·(-15)
Δ = 1185
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-\sqrt{1185}}{2*16}=\frac{-15-\sqrt{1185}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{1185}}{2*16}=\frac{-15+\sqrt{1185}}{32}
$