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1.8+16t-4.9t^2=0
a = -4.9; b = 16; c = +1.8;
Δ = b2-4ac
Δ = 162-4·(-4.9)·1.8
Δ = 291.28
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{-(16)-\sqrt{291.28}}{2*-4.9}=\frac{-16-\sqrt{291.28}}{-9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+\sqrt{291.28}}{2*-4.9}=\frac{-16+\sqrt{291.28}}{-9.8} $
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