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15z^2+23z+4=0
a = 15; b = 23; c = +4;
Δ = b2-4ac
Δ = 232-4·15·4
Δ = 289
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{289}=17$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-17}{2*15}=\frac{-40}{30} =-1+1/3 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+17}{2*15}=\frac{-6}{30} =-1/5 $
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