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50t+1.5t^2=84
We move all terms to the left:
50t+1.5t^2-(84)=0
a = 1.5; b = 50; c = -84;
Δ = b2-4ac
Δ = 502-4·1.5·(-84)
Δ = 3004
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3004}=\sqrt{4*751}=\sqrt{4}*\sqrt{751}=2\sqrt{751}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-2\sqrt{751}}{2*1.5}=\frac{-50-2\sqrt{751}}{3} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+2\sqrt{751}}{2*1.5}=\frac{-50+2\sqrt{751}}{3} $
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