f^2+16f=-9

Solution for f^2+16f=-9 equation:

f^2+16f=-9

We move all terms to the left:

f^2+16f-(-9)=0

We add all the numbers together, and all the variables

f^2+16f+9=0

a = 1; b = 16; c = +9;
Δ = b2-4ac
Δ = 162-4·1·9
Δ = 220
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:

$\sqrt{\Delta}=\sqrt{220}=\sqrt{4*55}=\sqrt{4}*\sqrt{55}=2\sqrt{55}$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{55}}{2*1}=\frac{-16-2\sqrt{55}}{2}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{55}}{2*1}=\frac{-16+2\sqrt{55}}{2}
$