How do you solve using the completing the square method #x^2+12x+23=0#?

We have that by applying the square method.

#x^2+12x+23=x^2+26x+6^2+(23-6^2)= (x+6)^2-13#

Hence from #x^2+12x+23=0=>(x+6)^2-13=0=> x+6=+-sqrt13=>x_1=-6+sqrt13 or x_2=-6-sqrt13#

To solve the equation ( x^2 + 12x + 23 = 0 ) using the completing the square method, follow these steps:

1. Move the constant term to the other side of the equation: ( x^2 + 12x = -23 )

2. Add and subtract the square of half of the coefficient of ( x ) (12/2 = 6): ( x^2 + 12x + 36 - 36 = -23 )

3. Factor the perfect square trinomial and simplify: ( (x + 6)^2 - 36 = -23 )

4. Add 36 to both sides of the equation to isolate the perfect square trinomial: ( (x + 6)^2 = 13 )

5. Take the square root of both sides: ( x + 6 = \pm \sqrt{13} )

6. Subtract 6 from both sides to solve for ( x ): ( x = -6 \pm \sqrt{13} )

Therefore, the solutions to the equation are ( x = -6 + \sqrt{13} ) and ( x = -6 - \sqrt{13} ).