125/3=w+w^2

Solution for 125/3=w+w^2 equation:

125/3=w+w^2

We move all terms to the left:

125/3-(w+w^2)=0

We get rid of parentheses

-w^2-w+125/3=0

We multiply all the terms by the denominator


-w^2*3-w*3+125=0

Wy multiply elements

-3w^2-3w+125=0

a = -3; b = -3; c = +125;
Δ = b2-4ac
Δ = -32-4·(-3)·125
Δ = 1509
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{-(-3)-\sqrt{1509}}{2*-3}=\frac{3-\sqrt{1509}}{-6}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{1509}}{2*-3}=\frac{3+\sqrt{1509}}{-6}
$