-5t^2+10t+2.5=0

Solution for -5t^2+10t+2.5=0 equation:

-5t^2+10t+2.5=0

a = -5; b = 10; c = +2.5;
Δ = b2-4ac
Δ = 102-4·(-5)·2.5
Δ = 150
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:

$\sqrt{\Delta}=\sqrt{150}=\sqrt{25*6}=\sqrt{25}*\sqrt{6}=5\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-5\sqrt{6}}{2*-5}=\frac{-10-5\sqrt{6}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+5\sqrt{6}}{2*-5}=\frac{-10+5\sqrt{6}}{-10}
$