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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

5t^2-5t-10=0

a = 5; b = -5; c = -10;
Δ = b2-4ac
Δ = -52-4·5·(-10)
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*5}=\frac{-10}{10}
=-1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+15}{2*5}=\frac{20}{10}
=2
$