Here's the question you clicked on:
chopsuey
Complete the square x^2+6x=7
\[x^2+6x-7=0\] \[(x+7)(x-1)=0\]
x^2+6x =7 remember x^2 = x^2+2ax+a^2 x^2 is x^2. 6x is 2ax, we have to find a we can find a by dividing 6x by 2, which is 3. So a =3 but remember a^2, so 3^2 is 9 (3 times 3 equals 9) now you can plug that in. x^2+6x+9=7 but what we do to one side we must do to the other x^2+6x+9=7+9 right off the bat we can add 7 and 9 x^2+6x+9=16 ok now, lets factor it out by using difference of two perfect squares (you can ignore 16 for now) (x+3)(x+3) now you can bring back the 16 (x+3)^2=16 btw (x+3)^2 is same as (x+3)(x+3) now, we have to find the square root of (x+3)^2 and 16 the square root of (x+3)^2 is just (x+3) and the square root of 16 is just 4 so (x+3) = (+-) 4 remember the plus or minus sign with the square roots now , subtract 3 from both sides x= -3 (+-) 4 now just do basic arithmetic x = -3 + 4 x= 1 is one solution x = -3 - 4 x= -7 is another solution
\[x^2+6x-7=0\]\[(x+3)^2-3^2-7=0\]\[(x+3)^2-16=0\]\[(x+3)^2=16\]\[x+3=\pm 4\]\[x=1 , -7\]