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6z^2+23z+7=0
a = 6; b = 23; c = +7;
Δ = b2-4ac
Δ = 232-4·6·7
Δ = 361
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{361}=19$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-19}{2*6}=\frac{-42}{12} =-3+1/2 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+19}{2*6}=\frac{-4}{12} =-1/3 $
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