15(t)=-16t^2+40

Solution for 15(t)=-16t^2+40 equation:

15(t)=-16t^2+40

We move all terms to the left:

15(t)-(-16t^2+40)=0

We get rid of parentheses

16t^2+15t-40=0

a = 16; b = 15; c = -40;
Δ = b2-4ac
Δ = 152-4·16·(-40)
Δ = 2785
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{2785}}{2*16}=\frac{-15-\sqrt{2785}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{2785}}{2*16}=\frac{-15+\sqrt{2785}}{32}
$