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z^2+24+14z=0
a = 1; b = 14; c = +24;
Δ = b2-4ac
Δ = 142-4·1·24
Δ = 100
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{100}=10$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-10}{2*1}=\frac{-24}{2} =-12 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+10}{2*1}=\frac{-4}{2} =-2 $
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