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15z^2+14z+3=0.
a = 15; b = 14; c = +3;
Δ = b2-4ac
Δ = 142-4·15·3
Δ = 16
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{16}=4$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-4}{2*15}=\frac{-18}{30} =-3/5 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+4}{2*15}=\frac{-10}{30} =-1/3 $
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