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-3t^2+60t=0
a = -3; b = 60; c = 0;
Δ = b2-4ac
Δ = 602-4·(-3)·0
Δ = 3600
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{3600}=60$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-60}{2*-3}=\frac{-120}{-6} =+20 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+60}{2*-3}=\frac{0}{-6} =0 $
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