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w^2+1w=63
We move all terms to the left:
w^2+1w-(63)=0
We add all the numbers together, and all the variables
w^2+w-63=0
a = 1; b = 1; c = -63;
Δ = b2-4ac
Δ = 12-4·1·(-63)
Δ = 253
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{253}}{2*1}=\frac{-1-\sqrt{253}}{2} $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{253}}{2*1}=\frac{-1+\sqrt{253}}{2} $
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