9n^2+4n+4=20

Solution for 9n^2+4n+4=20 equation:

9n^2+4n+4=20

We move all terms to the left:

9n^2+4n+4-(20)=0

We add all the numbers together, and all the variables

9n^2+4n-16=0

a = 9; b = 4; c = -16;
Δ = b2-4ac
Δ = 42-4·9·(-16)
Δ = 592
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{592}=\sqrt{16*37}=\sqrt{16}*\sqrt{37}=4\sqrt{37}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{37}}{2*9}=\frac{-4-4\sqrt{37}}{18}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{37}}{2*9}=\frac{-4+4\sqrt{37}}{18}
$