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(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} $
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