98t-16t^2=0

Solution for 98t-16t^2=0 equation:

98t-16t^2=0

a = -16; b = 98; c = 0;
Δ = b2-4ac
Δ = 982-4·(-16)·0
Δ = 9604
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9604}=98$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(98)-98}{2*-16}=\frac{-196}{-32}
=6+1/8
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(98)+98}{2*-16}=\frac{0}{-32}
=0
$