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60=2t+4.9t^2
We move all terms to the left:
60-(2t+4.9t^2)=0
We get rid of parentheses
-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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