w(3w+1)=5^2

Solution for w(3w+1)=5^2 equation:

w(3w+1)=5^2

We move all terms to the left:

w(3w+1)-(5^2)=0

We add all the numbers together, and all the variables

w(3w+1)-25=0

We multiply parentheses

3w^2+w-25=0

a = 3; b = 1; c = -25;
Δ = b2-4ac
Δ = 12-4·3·(-25)
Δ = 301
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{301}}{2*3}=\frac{-1-\sqrt{301}}{6}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{301}}{2*3}=\frac{-1+\sqrt{301}}{6}
$