Question

can someone solve x^2-6x=7

Answer 1

(x-7)(x+1)

Answer 2

\[x^2-6x-7 =0\rightarrow(x-7)(x+1)\rightarrow x=7, -1\]

Answer 3

King George gave you the right answer but to the procedure is as follows. First of all notice that we have a quadratic equation in the General Form\[ax ^{2}+bx+c=0\] where \[a, b, c\] are real constant coefficients or just simply constants. Now notice that to solve a quadratic equations is just another way to expressing the following: For what values of x does this equation equals zero? Therefore we are just find those values of x that gives zero when plugged in. There are 2 ways to solve this equation. The long one which is using the infamous quadratic equation:\[r = (-b \pm \sqrt{b ^{2}-4ac} ) /(2a)\] where r (r for roots) represent the values of x that makes the equation zero. And the shortest way which at first can be confusing but is much simple when practice:
Take \[a*c\] and find two factors of \[a*c\] that the sum equals \[b\]. In our example \[a =1\] and \[c =-7\] Notice that to have the equation in its general form we have to equal to zero therefore we have\[(1)x ^{2}+(-6)x+(-7)=0\]. Now \[a*c= (1)*(-7)\]. Now ask yourself, which two factors of \[(-7)\] the sum equals \[-6 =b\]? Well clearly \[1+(-7) = -6\] Therefore we have that the roots are \[(x +1)( x-7)=0\]. Now try using the quadratic equation to check:\[r = (-(-6)\pm \sqrt{(-6)^{2}-4(1)(-7)})/(2*1)\] which equals to\[r= (6\pm \sqrt{36+28)}/2\] which is\[r = (6\pm \sqrt{64})/2 = (6\pm8)/2\]. Now the plus and minus sign just makes to understand that we will have to different solutions:\[r _{1}=(6+8)/2 = 14/2 = 7\] and \[r _{2} = (6-8)/2=-2/2=-1\]. BUT WAIT WE HAVE ALTERED SIGNS. It's not an error, is just that the quadratic equation gives the exact roots, while the other form gives them as factors, which if you equate to zero both factors\[(x+1)(x-7)=0\] we have\[x+1 = 0 \rightarrow x=-1\] and \[x-7=0\rightarrow x=7\]. And there you go, beautiful as always. Peace, Love and Happiness from Puerto Rico