6z(4-z)=2z

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Solution for 6z(4-z)=2z equation:



6z(4-z)=2z
We move all terms to the left:
6z(4-z)-(2z)=0
We add all the numbers together, and all the variables
6z(-1z+4)-2z=0
We add all the numbers together, and all the variables
-2z+6z(-1z+4)=0
We multiply parentheses
-6z^2-2z+24z=0
We add all the numbers together, and all the variables
-6z^2+22z=0
a = -6; b = 22; c = 0;
Δ = b2-4ac
Δ = 222-4·(-6)·0
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-22}{2*-6}=\frac{-44}{-12} =3+2/3 $
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+22}{2*-6}=\frac{0}{-12} =0 $

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