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16t^2+135t=135
We move all terms to the left:
16t^2+135t-(135)=0
a = 16; b = 135; c = -135;
Δ = b2-4ac
Δ = 1352-4·16·(-135)
Δ = 26865
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{26865}=\sqrt{9*2985}=\sqrt{9}*\sqrt{2985}=3\sqrt{2985}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(135)-3\sqrt{2985}}{2*16}=\frac{-135-3\sqrt{2985}}{32} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(135)+3\sqrt{2985}}{2*16}=\frac{-135+3\sqrt{2985}}{32} $
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