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4.905x^2+10x-60=0
a = 4.905; b = 10; c = -60;
Δ = b2-4ac
Δ = 102-4·4.905·(-60)
Δ = 1277.2
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-\sqrt{1277.2}}{2*4.905}=\frac{-10-\sqrt{1277.2}}{9.81} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+\sqrt{1277.2}}{2*4.905}=\frac{-10+\sqrt{1277.2}}{9.81} $
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