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z^2+2z=12
We move all terms to the left:
z^2+2z-(12)=0
a = 1; b = 2; c = -12;
Δ = b2-4ac
Δ = 22-4·1·(-12)
Δ = 52
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{13}}{2*1}=\frac{-2-2\sqrt{13}}{2} $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{13}}{2*1}=\frac{-2+2\sqrt{13}}{2} $
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