7f^2+12f+4=0

Solution for 7f^2+12f+4=0 equation:

7f^2+12f+4=0

a = 7; b = 12; c = +4;
Δ = b2-4ac
Δ = 122-4·7·4
Δ = 32
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{2}}{2*7}=\frac{-12-4\sqrt{2}}{14}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{2}}{2*7}=\frac{-12+4\sqrt{2}}{14}
$