-4.9t^2+32t+124=0

Solution for -4.9t^2+32t+124=0 equation:

-4.9t^2+32t+124=0

a = -4.9; b = 32; c = +124;
Δ = b2-4ac
Δ = 322-4·(-4.9)·124
Δ = 3454.4
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{-(32)-\sqrt{3454.4}}{2*-4.9}=\frac{-32-\sqrt{3454.4}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+\sqrt{3454.4}}{2*-4.9}=\frac{-32+\sqrt{3454.4}}{-9.8}
$