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2w^2+w=36
We move all terms to the left:
2w^2+w-(36)=0
a = 2; b = 1; c = -36;
Δ = b2-4ac
Δ = 12-4·2·(-36)
Δ = 289
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}$$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{289}=17$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-17}{2*2}=\frac{-18}{4} =-4+1/2 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+17}{2*2}=\frac{16}{4} =4 $
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