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z^2-3z=+2
We move all terms to the left:
z^2-3z-(+2)=0
We add all the numbers together, and all the variables
z^2-3z-2=0
a = 1; b = -3; c = -2;
Δ = b2-4ac
Δ = -32-4·1·(-2)
Δ = 17
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}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{17}}{2*1}=\frac{3-\sqrt{17}}{2} $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{17}}{2*1}=\frac{3+\sqrt{17}}{2} $
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