18t+2t^2=180

Solution for 18t+2t^2=180 equation:

18t+2t^2=180

We move all terms to the left:

18t+2t^2-(180)=0

a = 2; b = 18; c = -180;
Δ = b2-4ac
Δ = 182-4·2·(-180)
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-42}{2*2}=\frac{-60}{4}
=-15
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+42}{2*2}=\frac{24}{4}
=6
$