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12z(1/6-18)=2z-3
We move all terms to the left:
12z(1/6-18)-(2z-3)=0
We multiply parentheses
12z^2-216z-(2z-3)=0
We get rid of parentheses
12z^2-216z-2z+3=0
We add all the numbers together, and all the variables
12z^2-218z+3=0
a = 12; b = -218; c = +3;
Δ = b2-4ac
Δ = -2182-4·12·3
Δ = 47380
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}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{47380}=\sqrt{4*11845}=\sqrt{4}*\sqrt{11845}=2\sqrt{11845}$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-218)-2\sqrt{11845}}{2*12}=\frac{218-2\sqrt{11845}}{24} $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-218)+2\sqrt{11845}}{2*12}=\frac{218+2\sqrt{11845}}{24} $
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