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4/3z=20z=

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Solution for 4/3z=20z= equation:



4/3z=20z=
We move all terms to the left:
4/3z-(20z)=0
Domain of the equation: 3z!=0
z!=0/3
z!=0
z∈R
We add all the numbers together, and all the variables
-20z+4/3z=0
We multiply all the terms by the denominator
-20z*3z+4=0
Wy multiply elements
-60z^2+4=0
a = -60; b = 0; c = +4;
Δ = b2-4ac
Δ = 02-4·(-60)·4
Δ = 960
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{960}=\sqrt{64*15}=\sqrt{64}*\sqrt{15}=8\sqrt{15}
z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{15}}{2*-60}=\frac{0-8\sqrt{15}}{-120} =-\frac{8\sqrt{15}}{-120} =-\frac{\sqrt{15}}{-15}
z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{15}}{2*-60}=\frac{0+8\sqrt{15}}{-120} =\frac{8\sqrt{15}}{-120} =\frac{\sqrt{15}}{-15}

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