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12z^2+13z+3=0
a = 12; b = 13; c = +3;
Δ = b2-4ac
Δ = 132-4·12·3
Δ = 25
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{25}=5$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-5}{2*12}=\frac{-18}{24} =-3/4 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+5}{2*12}=\frac{-8}{24} =-1/3 $
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