A word to the wise: Please use " ^ " to indicate exponentiation: x^2, not 2x.
x^2 + 6x = 7 can be re-written as x^2 + 6x = 7
Take HALF of the coefficient of x; in other words, take HALF of 6. Result: 3.
Square this result Add the square to x^2 + 6x, and then subtract it from
x^2 + 6x + 9: x^2 + 6x + 9 -9 = 7
Now rewrite x^2 + 6x + 9 in the form (x+3)^2 Equate that to 7+9 (or 16) on the right side.
Then you have (x+3)^2 = 16.
Take the square root of both sides: x+3 = sqrt(16) = plus or minus 4.
There are 2 roots: One is x = -3 + 4, or x = 1.
The other is x = -3 - 4, or x = -7
It's very important that you check your results. Is x =1 a solution?
Subst. 1 for x in x^2 + 6x = 7: 1^2 + 6(1) = 7? Yes, this is true, so x = 1 is a solution.
Subst. -7 for x in x^2 + 6x = 7: (-7)^2 + 6(-7) = 7 Is this true?
49 - 42 = 7? Yes, this is true, so x = -7 is another solution.
The "solution set" is {1, -7}.
