-13-26t+4.9t^2=0

Solution for -13-26t+4.9t^2=0 equation:

-13-26t+4.9t^2=0

a = 4.9; b = -26; c = -13;
Δ = b2-4ac
Δ = -262-4·4.9·(-13)
Δ = 930.8
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{-(-26)-\sqrt{930.8}}{2*4.9}=\frac{26-\sqrt{930.8}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+\sqrt{930.8}}{2*4.9}=\frac{26+\sqrt{930.8}}{9.8}
$