0=-5t^2+35t+150

Solution for 0=-5t^2+35t+150 equation:

0=-5t^2+35t+150

We move all terms to the left:

0-(-5t^2+35t+150)=0

We add all the numbers together, and all the variables

-(-5t^2+35t+150)=0

We get rid of parentheses

5t^2-35t-150=0

a = 5; b = -35; c = -150;
Δ = b2-4ac
Δ = -352-4·5·(-150)
Δ = 4225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4225}=65$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-65}{2*5}=\frac{-30}{10}
=-3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+65}{2*5}=\frac{100}{10}
=10
$