10=-4.9t^2+25t+13

Solution for 10=-4.9t^2+25t+13 equation:

10=-4.9t^2+25t+13

We move all terms to the left:

10-(-4.9t^2+25t+13)=0

We get rid of parentheses

4.9t^2-25t-13+10=0

We add all the numbers together, and all the variables

4.9t^2-25t-3=0

a = 4.9; b = -25; c = -3;
Δ = b2-4ac
Δ = -252-4·4.9·(-3)
Δ = 683.8
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{-(-25)-\sqrt{683.8}}{2*4.9}=\frac{25-\sqrt{683.8}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+\sqrt{683.8}}{2*4.9}=\frac{25+\sqrt{683.8}}{9.8}
$