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z2+6z-7=0
We add all the numbers together, and all the variables
z^2+6z-7=0
a = 1; b = 6; c = -7;
Δ = b2-4ac
Δ = 62-4·1·(-7)
Δ = 64
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{64}=8$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-8}{2*1}=\frac{-14}{2} =-7 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+8}{2*1}=\frac{2}{2} =1 $
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