4.9t^2+10t-50=0

Solution for 4.9t^2+10t-50=0 equation:

4.9t^2+10t-50=0

a = 4.9; b = 10; c = -50;
Δ = b2-4ac
Δ = 102-4·4.9·(-50)
Δ = 1080
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:

$\sqrt{\Delta}=\sqrt{1080}=\sqrt{36*30}=\sqrt{36}*\sqrt{30}=6\sqrt{30}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-6\sqrt{30}}{2*4.9}=\frac{-10-6\sqrt{30}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+6\sqrt{30}}{2*4.9}=\frac{-10+6\sqrt{30}}{9.8}
$