105(t)=-5t^2-10t+180

Solution for 105(t)=-5t^2-10t+180 equation:

105(t)=-5t^2-10t+180

We move all terms to the left:

105(t)-(-5t^2-10t+180)=0

We get rid of parentheses

5t^2+10t+105t-180=0

We add all the numbers together, and all the variables

5t^2+115t-180=0

a = 5; b = 115; c = -180;
Δ = b2-4ac
Δ = 1152-4·5·(-180)
Δ = 16825
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:

$\sqrt{\Delta}=\sqrt{16825}=\sqrt{25*673}=\sqrt{25}*\sqrt{673}=5\sqrt{673}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(115)-5\sqrt{673}}{2*5}=\frac{-115-5\sqrt{673}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(115)+5\sqrt{673}}{2*5}=\frac{-115+5\sqrt{673}}{10}
$