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1.3=4.9t^2
We move all terms to the left:
1.3-(4.9t^2)=0
We get rid of parentheses
-4.9t^2+1.3=0
a = -4.9; b = 0; c = +1.3;
Δ = b2-4ac
Δ = 02-4·(-4.9)·1.3
Δ = 25.48
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{-(0)-\sqrt{25.48}}{2*-4.9}=\frac{0-\sqrt{25.48}}{-9.8} =-\frac{\sqrt{}}{-9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{25.48}}{2*-4.9}=\frac{0+\sqrt{25.48}}{-9.8} =\frac{\sqrt{}}{-9.8} $
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