w^2+2w=w+3

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



w^2+2w=w+3
We move all terms to the left:
w^2+2w-(w+3)=0
We get rid of parentheses
w^2+2w-w-3=0
We add all the numbers together, and all the variables
w^2+w-3=0
a = 1; b = 1; c = -3;
Δ = b2-4ac
Δ = 12-4·1·(-3)
Δ = 13
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{13}}{2*1}=\frac{-1-\sqrt{13}}{2} $
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{13}}{2*1}=\frac{-1+\sqrt{13}}{2} $

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