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4.9t^2-19t-15=0
a = 4.9; b = -19; c = -15;
Δ = b2-4ac
Δ = -192-4·4.9·(-15)
Δ = 655
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{-(-19)-\sqrt{655}}{2*4.9}=\frac{19-\sqrt{655}}{9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{655}}{2*4.9}=\frac{19+\sqrt{655}}{9.8} $
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