Interesting method for solving quadratic equation. Help!!!!

[strugglewithmath93]

Question: An interesting method for solving quadratic equations came from India. The steps are
a) Move the constant term to the right side of the equation.
b) Multiply each term in the equation by four times the coefficient of the x² term.
c) Square the coefficient of the original x term and add it to both sides of the equation.
d) Take the square root of each side.
e) Set the left side of the equation equal to the positive square root of the number on the right side of the equation and solve for x.
f) Set the left side of the equation equal to the negative square root of the number on the right side of the equation and solve for x.
Example: Solve x²+3x-10=0
x² + 3x = 10
4x² + 12x = 40
4x² + 12x + 9 = 40 + 9
4x² + 12x + 9 = 49
2x + 3 = ±7
2x + 3 = 7
2x = 4
x = 2

2x + 3 = -7
2x = -10
x = -5

My first attempted problem is x² - 2x – 13 = 0
Step a) x² - 2x = 13
Step b) 4x² 8x = 52
Step c) 4x² 8x + 4 = 52 + 4
4x² 8x + 4 = 56
Step d) 2x – 2 = approximately ±7.4833
Step e) 2x – 2 = approx 7.4833
2x = approx 9.4833
x = approx 4.74165
Step f) 2x – 2 = approx -7.4833
2x = approx 5.4833
x = approx 2.74165

I feel I have done the problem wrong. Can you please help get me on the right track? Any help is very appreciated.

 

[wjm11]

Step f) 2x – 2 = approx -7.4833
2x = approx 5.4833
x = approx 2.74165

Your answers are correct except for a missing "-" sign. The last one should be -2.74 . (2x = approx -5.4833)