0=-5t2+10t+18

Solution for 0=-5t2+10t+18 equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

5t^2-10t-18=0

a = 5; b = -10; c = -18;
Δ = b2-4ac
Δ = -102-4·5·(-18)
Δ = 460
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:

$\sqrt{\Delta}=\sqrt{460}=\sqrt{4*115}=\sqrt{4}*\sqrt{115}=2\sqrt{115}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{115}}{2*5}=\frac{10-2\sqrt{115}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{115}}{2*5}=\frac{10+2\sqrt{115}}{10}
$