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4z^2+24z-364=0
a = 4; b = 24; c = -364;
Δ = b2-4ac
Δ = 242-4·4·(-364)
Δ = 6400
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{6400}=80$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-80}{2*4}=\frac{-104}{8} =-13 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+80}{2*4}=\frac{56}{8} =7 $
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