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4.9t^2+3t-30=0
a = 4.9; b = 3; c = -30;
Δ = b2-4ac
Δ = 32-4·4.9·(-30)
Δ = 597
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)-\sqrt{597}}{2*4.9}=\frac{-3-\sqrt{597}}{9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{597}}{2*4.9}=\frac{-3+\sqrt{597}}{9.8} $
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