0=-4.9t^2+50t+10

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

0=-4.9t^2+50t+10

We move all terms to the left:

0-(-4.9t^2+50t+10)=0

We add all the numbers together, and all the variables

-(-4.9t^2+50t+10)=0

We get rid of parentheses

4.9t^2-50t-10=0

a = 4.9; b = -50; c = -10;
Δ = b2-4ac
Δ = -502-4·4.9·(-10)
Δ = 2696
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:

$\sqrt{\Delta}=\sqrt{2696}=\sqrt{4*674}=\sqrt{4}*\sqrt{674}=2\sqrt{674}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-2\sqrt{674}}{2*4.9}=\frac{50-2\sqrt{674}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+2\sqrt{674}}{2*4.9}=\frac{50+2\sqrt{674}}{9.8}
$