3w+w2=7w-w2

Solution for 3w+w2=7w-w2 equation:

3w+w2=7w-w2

We move all terms to the left:

3w+w2-(7w-w2)=0

We add all the numbers together, and all the variables

-(+7w-1w^2)+3w+w2=0

We add all the numbers together, and all the variables

w^2-(+7w-1w^2)+3w=0

We get rid of parentheses

w^2+1w^2-7w+3w=0

We add all the numbers together, and all the variables

2w^2-4w=0

a = 2; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·2·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*2}=\frac{0}{4}
=0
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*2}=\frac{8}{4}
=2
$