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4t^2=108
We move all terms to the left:
4t^2-(108)=0
a = 4; b = 0; c = -108;
Δ = b2-4ac
Δ = 02-4·4·(-108)
Δ = 1728
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1728}=\sqrt{576*3}=\sqrt{576}*\sqrt{3}=24\sqrt{3}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24\sqrt{3}}{2*4}=\frac{0-24\sqrt{3}}{8} =-\frac{24\sqrt{3}}{8} =-3\sqrt{3} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24\sqrt{3}}{2*4}=\frac{0+24\sqrt{3}}{8} =\frac{24\sqrt{3}}{8} =3\sqrt{3} $
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