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4.9t^2+12t-21=0
a = 4.9; b = 12; c = -21;
Δ = b2-4ac
Δ = 122-4·4.9·(-21)
Δ = 555.6
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{-(12)-\sqrt{555.6}}{2*4.9}=\frac{-12-\sqrt{555.6}}{9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+\sqrt{555.6}}{2*4.9}=\frac{-12+\sqrt{555.6}}{9.8} $
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