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12z^2+17z+6=0.
a = 12; b = 17; c = +6;
Δ = b2-4ac
Δ = 172-4·12·6
Δ = 1
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{1}=1$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-1}{2*12}=\frac{-18}{24} =-3/4 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+1}{2*12}=\frac{-16}{24} =-2/3 $
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