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180=3t^2
We move all terms to the left:
180-(3t^2)=0
a = -3; b = 0; c = +180;
Δ = b2-4ac
Δ = 02-4·(-3)·180
Δ = 2160
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2160}=\sqrt{144*15}=\sqrt{144}*\sqrt{15}=12\sqrt{15}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{15}}{2*-3}=\frac{0-12\sqrt{15}}{-6} =-\frac{12\sqrt{15}}{-6} =-\frac{2\sqrt{15}}{-1} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{15}}{2*-3}=\frac{0+12\sqrt{15}}{-6} =\frac{12\sqrt{15}}{-6} =\frac{2\sqrt{15}}{-1} $
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