2w(w+3w)=41

Solution for 2w(w+3w)=41 equation:

2w(w+3w)=41

We move all terms to the left:

2w(w+3w)-(41)=0

We add all the numbers together, and all the variables

2w(+4w)-41=0

We multiply parentheses

8w^2-41=0

a = 8; b = 0; c = -41;
Δ = b2-4ac
Δ = 02-4·8·(-41)
Δ = 1312
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:

$\sqrt{\Delta}=\sqrt{1312}=\sqrt{16*82}=\sqrt{16}*\sqrt{82}=4\sqrt{82}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{82}}{2*8}=\frac{0-4\sqrt{82}}{16}
=-\frac{4\sqrt{82}}{16}
=-\frac{\sqrt{82}}{4}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{82}}{2*8}=\frac{0+4\sqrt{82}}{16}
=\frac{4\sqrt{82}}{16}
=\frac{\sqrt{82}}{4}
$