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-4.9t^2+30.5t+9.9=0
a = -4.9; b = 30.5; c = +9.9;
Δ = b2-4ac
Δ = 30.52-4·(-4.9)·9.9
Δ = 1124.29
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{-(30.5)-\sqrt{1124.29}}{2*-4.9}=\frac{-30.5-\sqrt{1124.29}}{-9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30.5)+\sqrt{1124.29}}{2*-4.9}=\frac{-30.5+\sqrt{1124.29}}{-9.8} $
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