-5t^2+10t+18=18

Solution for -5t^2+10t+18=18 equation:

-5t^2+10t+18=18

We move all terms to the left:

-5t^2+10t+18-(18)=0

We add all the numbers together, and all the variables

-5t^2+10t=0

a = -5; b = 10; c = 0;
Δ = b2-4ac
Δ = 102-4·(-5)·0
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10}{2*-5}=\frac{-20}{-10}
=+2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10}{2*-5}=\frac{0}{-10}
=0
$