w(w-6)+w(w-6)=1

Solution for w(w-6)+w(w-6)=1 equation:

w(w-6)+w(w-6)=1

We move all terms to the left:

w(w-6)+w(w-6)-(1)=0

We multiply parentheses

w^2+w^2-6w-6w-1=0

We add all the numbers together, and all the variables

2w^2-12w-1=0

a = 2; b = -12; c = -1;
Δ = b2-4ac
Δ = -122-4·2·(-1)
Δ = 152
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:

$\sqrt{\Delta}=\sqrt{152}=\sqrt{4*38}=\sqrt{4}*\sqrt{38}=2\sqrt{38}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{38}}{2*2}=\frac{12-2\sqrt{38}}{4}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{38}}{2*2}=\frac{12+2\sqrt{38}}{4}
$