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t=-16t^2+162
We move all terms to the left:
t-(-16t^2+162)=0
We get rid of parentheses
16t^2+t-162=0
a = 16; b = 1; c = -162;
Δ = b2-4ac
Δ = 12-4·16·(-162)
Δ = 10369
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{-(1)-\sqrt{10369}}{2*16}=\frac{-1-\sqrt{10369}}{32} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{10369}}{2*16}=\frac{-1+\sqrt{10369}}{32} $
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