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-4.9t^2+3.5t+5=0
a = -4.9; b = 3.5; c = +5;
Δ = b2-4ac
Δ = 3.52-4·(-4.9)·5
Δ = 110.25
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{-(3.5)-\sqrt{110.25}}{2*-4.9}=\frac{-3.5-\sqrt{110.25}}{-9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3.5)+\sqrt{110.25}}{2*-4.9}=\frac{-3.5+\sqrt{110.25}}{-9.8} $
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