(2x^2-5x)/(3x-7)=3/2

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Solution for (2x^2-5x)/(3x-7)=3/2 equation: 


D( x )

3*x-7 = 0

3*x-7 = 0

3*x-7 = 0

3*x-7 = 0 // + 7

3*x = 7 // : 3

x = 7/3

x in (-oo:7/3) U (7/3:+oo)

(2*x^2-(5*x))/(3*x-7) = 3/2 // - 3/2

(2*x^2-(5*x))/(3*x-7)-(3/2) = 0

(2*x^2-5*x)/(3*x-7)-3/2 = 0

(2*(2*x^2-5*x))/(2*(3*x-7))+(-3*(3*x-7))/(2*(3*x-7)) = 0

2*(2*x^2-5*x)-3*(3*x-7) = 0

4*x^2-19*x+21 = 0

4*x^2-19*x+21 = 0

4*x^2-19*x+21 = 0

DELTA = (-19)^2-(4*4*21)

DELTA = 25

DELTA > 0

x = (25^(1/2)+19)/(2*4) or x = (19-25^(1/2))/(2*4)

x = 3 or x = 7/4

(x-7/4)*(x-3) = 0

((x-7/4)*(x-3))/(2*(3*x-7)) = 0

((x-7/4)*(x-3))/(2*(3*x-7)) = 0 // * 2*(3*x-7)

(x-7/4)*(x-3) = 0

( x-3 )

x-3 = 0 // + 3

x = 3

( x-7/4 )

x-7/4 = 0 // + 7/4

x = 7/4

x in { 3, 7/4 }

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