10=17t+4.9t^2

Solution for 10=17t+4.9t^2 equation:

10=17t+4.9t^2

We move all terms to the left:

10-(17t+4.9t^2)=0

We get rid of parentheses

-4.9t^2-17t+10=0

a = -4.9; b = -17; c = +10;
Δ = b2-4ac
Δ = -172-4·(-4.9)·10
Δ = 485
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{-(-17)-\sqrt{485}}{2*-4.9}=\frac{17-\sqrt{485}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+\sqrt{485}}{2*-4.9}=\frac{17+\sqrt{485}}{-9.8}
$