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w^2+4w=2
We move all terms to the left:
w^2+4w-(2)=0
a = 1; b = 4; c = -2;
Δ = b2-4ac
Δ = 42-4·1·(-2)
Δ = 24
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}$$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{6}}{2*1}=\frac{-4-2\sqrt{6}}{2} $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{6}}{2*1}=\frac{-4+2\sqrt{6}}{2} $
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