-2t^2+10t+100=0

Solution for -2t^2+10t+100=0 equation:

-2t^2+10t+100=0

a = -2; b = 10; c = +100;
Δ = b2-4ac
Δ = 102-4·(-2)·100
Δ = 900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{900}=30$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-30}{2*-2}=\frac{-40}{-4}
=+10
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+30}{2*-2}=\frac{20}{-4}
=-5
$