2f(8f+3)=0

Solution for 2f(8f+3)=0 equation:

2f(8f+3)=0

We multiply parentheses

16f^2+6f=0

a = 16; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·16·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*16}=\frac{-12}{32}
=-3/8
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*16}=\frac{0}{32}
=0
$