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

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

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

a = -10; b = 5; c = +10;
Δ = b2-4ac
Δ = 52-4·(-10)·10
Δ = 425
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:

$\sqrt{\Delta}=\sqrt{425}=\sqrt{25*17}=\sqrt{25}*\sqrt{17}=5\sqrt{17}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5\sqrt{17}}{2*-10}=\frac{-5-5\sqrt{17}}{-20}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5\sqrt{17}}{2*-10}=\frac{-5+5\sqrt{17}}{-20}
$