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3t^2=39
We move all terms to the left:
3t^2-(39)=0
a = 3; b = 0; c = -39;
Δ = b2-4ac
Δ = 02-4·3·(-39)
Δ = 468
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{468}=\sqrt{36*13}=\sqrt{36}*\sqrt{13}=6\sqrt{13}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{13}}{2*3}=\frac{0-6\sqrt{13}}{6} =-\frac{6\sqrt{13}}{6} =-\sqrt{13} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{13}}{2*3}=\frac{0+6\sqrt{13}}{6} =\frac{6\sqrt{13}}{6} =\sqrt{13} $
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