32f^2+98f=0

Solution for 32f^2+98f=0 equation:

32f^2+98f=0

a = 32; b = 98; c = 0;
Δ = b2-4ac
Δ = 982-4·32·0
Δ = 9604
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9604}=98$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(98)-98}{2*32}=\frac{-196}{64}
=-3+1/16
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(98)+98}{2*32}=\frac{0}{64}
=0
$