25t-5t^2=15

Solution for 25t-5t^2=15 equation:

25t-5t^2=15

We move all terms to the left:

25t-5t^2-(15)=0

a = -5; b = 25; c = -15;
Δ = b2-4ac
Δ = 252-4·(-5)·(-15)
Δ = 325
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:

$\sqrt{\Delta}=\sqrt{325}=\sqrt{25*13}=\sqrt{25}*\sqrt{13}=5\sqrt{13}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-5\sqrt{13}}{2*-5}=\frac{-25-5\sqrt{13}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+5\sqrt{13}}{2*-5}=\frac{-25+5\sqrt{13}}{-10}
$