(t)=-162t^2+32t+48

Solution for (t)=-162t^2+32t+48 equation:

(t)=-162t^2+32t+48

We move all terms to the left:

(t)-(-162t^2+32t+48)=0

We get rid of parentheses

162t^2-32t+t-48=0

We add all the numbers together, and all the variables

162t^2-31t-48=0

a = 162; b = -31; c = -48;
Δ = b2-4ac
Δ = -312-4·162·(-48)
Δ = 32065
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:

$\sqrt{\Delta}=\sqrt{32065}=\sqrt{121*265}=\sqrt{121}*\sqrt{265}=11\sqrt{265}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-31)-11\sqrt{265}}{2*162}=\frac{31-11\sqrt{265}}{324}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+11\sqrt{265}}{2*162}=\frac{31+11\sqrt{265}}{324}
$