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f(x)=x^2+6x-7
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Remember: \[ax^2+bx+c=0\] If what you are asking is how to solve: \[x^2+6x-7=0\] then you can complete the square: Start by adding 7 to both sides so you get: \[x^2+6x=7\] Divide "b-term" by 2 and then square it and add it to both sides so: \[x^2+6x+9=16\] (basically i am doing half of 6 and squaring it and adding it to both sides. then factor the left side by doing (x+b-term/2)(x+b-term/2): (x+3)(x+3)=16 becomes: \[(x+3)^2=16\] take the square root of both sides: x+3= -4 or x+3 = 4 subtract 3 from both sides: x=1 or x=−7
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