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12z^2+19z+4=0
a = 12; b = 19; c = +4;
Δ = b2-4ac
Δ = 192-4·12·4
Δ = 169
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{169}=13$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-13}{2*12}=\frac{-32}{24} =-1+1/3 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+13}{2*12}=\frac{-6}{24} =-1/4 $
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