180=67(t)+0.5(t^2)

Solution for 180=67(t)+0.5(t^2) equation:

180=67(t)+0.5(t^2)

We move all terms to the left:

180-(67(t)+0.5(t^2))=0

We get rid of parentheses

-0.5t^2-67t+180=0

a = -0.5; b = -67; c = +180;
Δ = b2-4ac
Δ = -672-4·(-0.5)·180
Δ = 4849
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-67)-\sqrt{4849}}{2*-0.5}=\frac{67-\sqrt{4849}}{-1}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-67)+\sqrt{4849}}{2*-0.5}=\frac{67+\sqrt{4849}}{-1}
$