4.9t^2-22t+15=0

Solution for 4.9t^2-22t+15=0 equation:

4.9t^2-22t+15=0

a = 4.9; b = -22; c = +15;
Δ = b2-4ac
Δ = -222-4·4.9·15
Δ = 190
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)-\sqrt{190}}{2*4.9}=\frac{22-\sqrt{190}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+\sqrt{190}}{2*4.9}=\frac{22+\sqrt{190}}{9.8}
$