(2w-5)w=230

Solution for (2w-5)w=230 equation:

(2w-5)w=230

We move all terms to the left:

(2w-5)w-(230)=0

We multiply parentheses

2w^2-5w-230=0

a = 2; b = -5; c = -230;
Δ = b2-4ac
Δ = -52-4·2·(-230)
Δ = 1865
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{-(-5)-\sqrt{1865}}{2*2}=\frac{5-\sqrt{1865}}{4}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{1865}}{2*2}=\frac{5+\sqrt{1865}}{4}
$