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3t^2-6t-21=0
a = 3; b = -6; c = -21;
Δ = b2-4ac
Δ = -62-4·3·(-21)
Δ = 288
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-12\sqrt{2}}{2*3}=\frac{6-12\sqrt{2}}{6} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+12\sqrt{2}}{2*3}=\frac{6+12\sqrt{2}}{6} $
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