60+20(t)=2t^2

Solution for 60+20(t)=2t^2 equation:

60+20(t)=2t^2

We move all terms to the left:

60+20(t)-(2t^2)=0

determiningTheFunctionDomain
-2t^2+20t+60=0

a = -2; b = 20; c = +60;
Δ = b2-4ac
Δ = 202-4·(-2)·60
Δ = 880
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:

$\sqrt{\Delta}=\sqrt{880}=\sqrt{16*55}=\sqrt{16}*\sqrt{55}=4\sqrt{55}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{55}}{2*-2}=\frac{-20-4\sqrt{55}}{-4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{55}}{2*-2}=\frac{-20+4\sqrt{55}}{-4}
$