-4.9t^2+15t-5=0

Solution for -4.9t^2+15t-5=0 equation:

-4.9t^2+15t-5=0

a = -4.9; b = 15; c = -5;
Δ = b2-4ac
Δ = 152-4·(-4.9)·(-5)
Δ = 127
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{-(15)-\sqrt{127}}{2*-4.9}=\frac{-15-\sqrt{127}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{127}}{2*-4.9}=\frac{-15+\sqrt{127}}{-9.8}
$