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4z^2+4z-15=0
a = 4; b = 4; c = -15;
Δ = b2-4ac
Δ = 42-4·4·(-15)
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$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-16}{2*4}=\frac{-20}{8} =-2+1/2 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+16}{2*4}=\frac{12}{8} =1+1/2 $
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