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z2+z-6=0
We add all the numbers together, and all the variables
z^2+z-6=0
a = 1; b = 1; c = -6;
Δ = b2-4ac
Δ = 12-4·1·(-6)
Δ = 25
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{25}=5$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-5}{2*1}=\frac{-6}{2} =-3 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+5}{2*1}=\frac{4}{2} =2 $
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