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-4.9t^2-5t+150=0
a = -4.9; b = -5; c = +150;
Δ = b2-4ac
Δ = -52-4·(-4.9)·150
Δ = 2965
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{2965}}{2*-4.9}=\frac{5-\sqrt{2965}}{-9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{2965}}{2*-4.9}=\frac{5+\sqrt{2965}}{-9.8} $
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