20=-4f(f+6)+14

Solution for 20=-4f(f+6)+14 equation:

20=-4f(f+6)+14

We move all terms to the left:

20-(-4f(f+6)+14)=0


We calculate terms in parentheses: -(-4f(f+6)+14), so:

-4f(f+6)+14

We multiply parentheses

-4f^2-24f+14

Back to the equation:

-(-4f^2-24f+14)


We get rid of parentheses

4f^2+24f-14+20=0

We add all the numbers together, and all the variables

4f^2+24f+6=0

a = 4; b = 24; c = +6;
Δ = b2-4ac
Δ = 242-4·4·6
Δ = 480
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:

$\sqrt{\Delta}=\sqrt{480}=\sqrt{16*30}=\sqrt{16}*\sqrt{30}=4\sqrt{30}$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-4\sqrt{30}}{2*4}=\frac{-24-4\sqrt{30}}{8}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+4\sqrt{30}}{2*4}=\frac{-24+4\sqrt{30}}{8}
$