10=10t+2.5t^2

Solution for 10=10t+2.5t^2 equation:

10=10t+2.5t^2

We move all terms to the left:

10-(10t+2.5t^2)=0

We get rid of parentheses

-2.5t^2-10t+10=0

a = -2.5; b = -10; c = +10;
Δ = b2-4ac
Δ = -102-4·(-2.5)·10
Δ = 200
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:

$\sqrt{\Delta}=\sqrt{200}=\sqrt{100*2}=\sqrt{100}*\sqrt{2}=10\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-10\sqrt{2}}{2*-2.5}=\frac{10-10\sqrt{2}}{-5}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+10\sqrt{2}}{2*-2.5}=\frac{10+10\sqrt{2}}{-5}
$