2(4w*3w)+2(w*4w)=270

Solution for 2(4w*3w)+2(w*4w)=270 equation:

2(4w*3w)+2(w*4w)=270

We move all terms to the left:

2(4w*3w)+2(w*4w)-(270)=0

We add all the numbers together, and all the variables

2(+4w*3w)+2(+w*4w)-270=0

We multiply parentheses

24w^2+8w^2-270=0

We add all the numbers together, and all the variables

32w^2-270=0

a = 32; b = 0; c = -270;
Δ = b2-4ac
Δ = 02-4·32·(-270)
Δ = 34560
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:

$\sqrt{\Delta}=\sqrt{34560}=\sqrt{2304*15}=\sqrt{2304}*\sqrt{15}=48\sqrt{15}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-48\sqrt{15}}{2*32}=\frac{0-48\sqrt{15}}{64}
=-\frac{48\sqrt{15}}{64}
=-\frac{3\sqrt{15}}{4}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+48\sqrt{15}}{2*32}=\frac{0+48\sqrt{15}}{64}
=\frac{48\sqrt{15}}{64}
=\frac{3\sqrt{15}}{4}
$