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z^2=8z-12
We move all terms to the left:
z^2-(8z-12)=0
We get rid of parentheses
z^2-8z+12=0
a = 1; b = -8; c = +12;
Δ = b2-4ac
Δ = -82-4·1·12
Δ = 16
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}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{16}=4$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4}{2*1}=\frac{4}{2} =2 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4}{2*1}=\frac{12}{2} =6 $
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