0=3/4(x-5)^2-12

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Solution for 0=3/4(x-5)^2-12 equation: 


x in (-oo:+oo)

0 = (3/4)*(x-5)^2-12 // - (3/4)*(x-5)^2-12

0-((3/4)*(x-5)^2)+12 = 0

(-3/4)*(x-5)^2+0+12 = 0

0-3/4*(x-5)^2+12 = 0

0-3/4*(x-5)^2+12 = 0

15/2*x-3/4*x^2-75/4+12 = 0

15/2*x-3/4*x^2-27/4 = 0

15/2*x-3/4*x^2-27/4 = 0

3/2*(5*x-1/2*x^2-9/2) = 0

5*x-1/2*x^2-9/2 = 0

DELTA = 5^2-(-9/2*(-1/2)*4)

DELTA = 16

DELTA > 0

x = (16^(1/2)-5)/(-1/2*2) or x = (-16^(1/2)-5)/(-1/2*2)

x = 1 or x = 9

3/2*(x-1)*(x-9) = 0

3/2*(x-1)*(x-9) = 0

( 3/2 )

3/2 = 0

x belongs to the empty set

( x-1 )

x-1 = 0 // + 1

x = 1

( x-9 )

x-9 = 0 // + 9

x = 9

x in { 1, 9 }

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