(2x-3)^2/5=45

Solution for (2x-3)^2/5=45 equation:

x in (-oo:+oo)

((2*x-3)^2)/5 = 45 // - 45

((2*x-3)^2)/5-45 = 0

((2*x-3)^2)/5+(-45*5)/5 = 0

(2*x-3)^2-45*5 = 0

4*x^2-12*x-216 = 0

4*x^2-12*x-216 = 0

4*(x^2-3*x-54) = 0

x^2-3*x-54 = 0

DELTA = (-3)^2-(-54*1*4)

DELTA = 225

DELTA > 0

x = (225^(1/2)+3)/(1*2) or x = (3-225^(1/2))/(1*2)

x = 9 or x = -6

4*(x+6)*(x-9) = 0

(4*(x+6)*(x-9))/5 = 0

(4*(x+6)*(x-9))/5 = 0 // * 5

4*(x+6)*(x-9) = 0

( 4 )

4 = 0

x belongs to the empty set

( x+6 )

x+6 = 0 // - 6

x = -6

( x-9 )

x-9 = 0 // + 9

x = 9

x in { -6, 9 }