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z^2+5z=14
We move all terms to the left:
z^2+5z-(14)=0
a = 1; b = 5; c = -14;
Δ = b2-4ac
Δ = 52-4·1·(-14)
Δ = 81
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{81}=9$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-9}{2*1}=\frac{-14}{2} =-7 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+9}{2*1}=\frac{4}{2} =2 $
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