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t=-4.9t^2+12
We move all terms to the left:
t-(-4.9t^2+12)=0
We get rid of parentheses
4.9t^2+t-12=0
a = 4.9; b = 1; c = -12;
Δ = b2-4ac
Δ = 12-4·4.9·(-12)
Δ = 236.2
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{-(1)-\sqrt{236.2}}{2*4.9}=\frac{-1-\sqrt{236.2}}{9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{236.2}}{2*4.9}=\frac{-1+\sqrt{236.2}}{9.8} $
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