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c^2+4c=-2
We move all terms to the left:
c^2+4c-(-2)=0
We add all the numbers together, and all the variables
c^2+4c+2=0
a = 1; b = 4; c = +2;
Δ = b2-4ac
Δ = 42-4·1·2
Δ = 8
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{2}}{2*1}=\frac{-4-2\sqrt{2}}{2} $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{2}}{2*1}=\frac{-4+2\sqrt{2}}{2} $
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