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2z^2+12z-14=0
a = 2; b = 12; c = -14;
Δ = b2-4ac
Δ = 122-4·2·(-14)
Δ = 256
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{256}=16$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-16}{2*2}=\frac{-28}{4} =-7 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+16}{2*2}=\frac{4}{4} =1 $
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