4w^2+12w+1=121

Solution for 4w^2+12w+1=121 equation:

4w^2+12w+1=121

We move all terms to the left:

4w^2+12w+1-(121)=0

We add all the numbers together, and all the variables

4w^2+12w-120=0

a = 4; b = 12; c = -120;
Δ = b2-4ac
Δ = 122-4·4·(-120)
Δ = 2064
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2064}=\sqrt{16*129}=\sqrt{16}*\sqrt{129}=4\sqrt{129}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{129}}{2*4}=\frac{-12-4\sqrt{129}}{8}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{129}}{2*4}=\frac{-12+4\sqrt{129}}{8}
$