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4z^2+14z-30=0
a = 4; b = 14; c = -30;
Δ = b2-4ac
Δ = 142-4·4·(-30)
Δ = 676
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{676}=26$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-26}{2*4}=\frac{-40}{8} =-5 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+26}{2*4}=\frac{12}{8} =1+1/2 $
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