4.9t^2-28t+15=0

Solution for 4.9t^2-28t+15=0 equation:

4.9t^2-28t+15=0

a = 4.9; b = -28; c = +15;
Δ = b2-4ac
Δ = -282-4·4.9·15
Δ = 490
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{490}=\sqrt{49*10}=\sqrt{49}*\sqrt{10}=7\sqrt{10}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-7\sqrt{10}}{2*4.9}=\frac{28-7\sqrt{10}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+7\sqrt{10}}{2*4.9}=\frac{28+7\sqrt{10}}{9.8}
$