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

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

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

a = -10; b = 5; c = +5;
Δ = b2-4ac
Δ = 52-4·(-10)·5
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-15}{2*-10}=\frac{-20}{-20}
=1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+15}{2*-10}=\frac{10}{-20}
=-1/2
$