(2/3)(4x-5)=8

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Solution for (2/3)(4x-5)=8 equation: 



(2/3)(4x-5)=8

We move all terms to the left:

(2/3)(4x-5)-(8)=0



Domain of the equation: 3)(4x-5)!=0

x∈R



We add all the numbers together, and all the variables

(+2/3)(4x-5)-8=0

We multiply parentheses ..

(+8x^2+2/3*-5)-8=0

We multiply all the terms by the denominator


(+8x^2+2-8*3*-5)=0

We get rid of parentheses

8x^2+2-5-8*3*=0

We add all the numbers together, and all the variables

8x^2=0

a = 8; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·8·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{0}{16}=0$

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