-4.9t^2+19.05t-3.6=0

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Solution for -4.9t^2+19.05t-3.6=0 equation: 



-4.9t^2+19.05t-3.6=0

a = -4.9; b = 19.05; c = -3.6;
Δ = b2-4ac
Δ = 19.052-4·(-4.9)·(-3.6)
Δ = 292.3425
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{-(19.05)-\sqrt{292.3425}}{2*-4.9}=\frac{-19.05-\sqrt{292.3425}}{-9.8}
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19.05)+\sqrt{292.3425}}{2*-4.9}=\frac{-19.05+\sqrt{292.3425}}{-9.8}
$

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