First, put the equation in standard quadratic form:
(x - 3)^2 = 5(x−3)2=5
(x - 3)(x - 3) = 5(x−3)(x−3)=5
x^2 - 3x - 3x + 9 = 5x2−3x−3x+9=5
x^2 - 6x + 9 = 5x2−6x+9=5
x^2 - 6x + 9 - color(red)(5) = 5 - color(red)(5)x2−6x+9−5=5−5
x^2 - 6x + 4 = 0x2−6x+4=0
We can now use the quadratic equation to solve this problem:
The quadratic formula states:
For color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0ax2+bx+c=0, the values of xx which are the solutions to the equation are given by:
x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))x=−b±√b2−(4ac)2⋅a
Substituting:
color(red)(1)1 for color(red)(a)a
color(blue)(-6)−6 for color(blue)(b)b
color(green)(4)4 for color(green)(c)c gives:
x = (-color(blue)(-6) +- sqrt(color(blue)(-6)^2 - (4 * color(red)(1) * color(green)(4))))/(2 * color(red)(1))x=−−6±√−62−(4⋅1⋅4)2⋅1
x = (6 +- sqrt(36 - 16))/2x=6±√36−162
x = (6 +- sqrt(20))/2x=6±√202
x = (6 - sqrt(4 * 5))/2x=6−√4⋅52; x = (6 + sqrt(4 * 5))/2x=6+√4⋅52
x = (6 - sqrt(4)sqrt(5))/2x=6−√4√52; x = (6 + sqrt(4)sqrt(5))/2x=6+√4√52
x = (6 - 2sqrt(5))/2x=6−2√52; x = (6 + 2sqrt(5))/2x=6+2√52
x = 6/2 - (2sqrt(5))/2x=62−2√52; x = 6/2 + (2sqrt(5))/2x=62+2√52
x = 3 - sqrt(5)x=3−√5; x = 3 + sqrt(5)x=3+√5
