How do you solve #(x + 2)^2 = 13#?

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Answer 1 · 2018-03-11T14:09:05

#color(red)(x = -2 + sqrt13, color(blue)( -2 - sqrt13#

#(x + 2)^2 = 13#

#x^2 + 4x + 4 - 13 = 0#

#x^2 + 4x - 9 = 0#

Standard form of quadratic equation is

#ax^2 + bx + c = 0# and it’s roots are

# x = (-b +- sqrt (b^2 - 4ac)) / 2a#

For the given equation,

#x = (-4 +- sqrt(4^2 + (4*1*9)))/ (2 * 1)#

#x = (-4 +- sqrt(52)) / 2 = (-2 +- sqrt13)#

#color(red)(x = -2 + sqrt13, color(blue)( -2 - sqrt13#