w^2+2w+1=4w^2-14+6

Solution for w^2+2w+1=4w^2-14+6 equation:

w^2+2w+1=4w^2-14+6

We move all terms to the left:

w^2+2w+1-(4w^2-14+6)=0

We get rid of parentheses

w^2-4w^2+2w+14-6+1=0

We add all the numbers together, and all the variables

-3w^2+2w+9=0

a = -3; b = 2; c = +9;
Δ = b2-4ac
Δ = 22-4·(-3)·9
Δ = 112
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:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4\sqrt{7}}{2*-3}=\frac{-2-4\sqrt{7}}{-6}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4\sqrt{7}}{2*-3}=\frac{-2+4\sqrt{7}}{-6}
$