-4.9*t^2+26t=0

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

-4.9t^2+26t=0

a = -4.9; b = 26; c = 0;
Δ = b2-4ac
Δ = 262-4·(-4.9)·0
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-26}{2*-4.9}=\frac{-52}{-9.8}
=5+1/3.2666666666667
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+26}{2*-4.9}=\frac{0}{-9.8}
=0
$