15=40t-5t2

Solution for 15=40t-5t2 equation:

15=40t-5t^2

We move all terms to the left:

15-(40t-5t^2)=0

We get rid of parentheses

5t^2-40t+15=0

a = 5; b = -40; c = +15;
Δ = b2-4ac
Δ = -402-4·5·15
Δ = 1300
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:

$\sqrt{\Delta}=\sqrt{1300}=\sqrt{100*13}=\sqrt{100}*\sqrt{13}=10\sqrt{13}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-10\sqrt{13}}{2*5}=\frac{40-10\sqrt{13}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+10\sqrt{13}}{2*5}=\frac{40+10\sqrt{13}}{10}
$