-4.9t^2+22.5t+2=0

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Solution for -4.9t^2+22.5t+2=0 equation:



-4.9t^2+22.5t+2=0
a = -4.9; b = 22.5; c = +2;
Δ = b2-4ac
Δ = 22.52-4·(-4.9)·2
Δ = 545.45
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{-(22.5)-\sqrt{545.45}}{2*-4.9}=\frac{-22.5-\sqrt{545.45}}{-9.8} $
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22.5)+\sqrt{545.45}}{2*-4.9}=\frac{-22.5+\sqrt{545.45}}{-9.8} $

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