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4.9t^2+2t-60=0
a = 4.9; b = 2; c = -60;
Δ = b2-4ac
Δ = 22-4·4.9·(-60)
Δ = 1180
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1180}=\sqrt{4*295}=\sqrt{4}*\sqrt{295}=2\sqrt{295}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{295}}{2*4.9}=\frac{-2-2\sqrt{295}}{9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{295}}{2*4.9}=\frac{-2+2\sqrt{295}}{9.8} $
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