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60+20t-5t^2=0
a = -5; b = 20; c = +60;
Δ = b2-4ac
Δ = 202-4·(-5)·60
Δ = 1600
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{1600}=40$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-40}{2*-5}=\frac{-60}{-10} =+6 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+40}{2*-5}=\frac{20}{-10} =-2 $
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