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3t^2-24t-36=0
a = 3; b = -24; c = -36;
Δ = b2-4ac
Δ = -242-4·3·(-36)
Δ = 1008
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1008}=\sqrt{144*7}=\sqrt{144}*\sqrt{7}=12\sqrt{7}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-12\sqrt{7}}{2*3}=\frac{24-12\sqrt{7}}{6} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+12\sqrt{7}}{2*3}=\frac{24+12\sqrt{7}}{6} $
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