How do you complete the square for $x^{2} + 6 x - 7 = 0$ $x^2+6x-7=0$ ?

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Answer 1 · 2018-07-10T06:23:52

${(x + 3)}^{2} = 16$ $(x+3)^2=16$

The given equation:

$x^{2} + 6 x - 7 = 0$ $x^2+6x-7=0$

$x^{2} + 2 \cdot 3 \cdot x - 7 = 0$ $x^2+2\cdot 3\cdotx-7=0$

$x^{2} + 2 \cdot 3 \cdot x + 3^{2} - 3^{3} - 7 = 0$ $x^2+2\cdot 3\cdot x+3^2-3^3-7=0$

${(x + 3)}^{2} - 16 = 0$ $(x+3)^2-16=0$

${(x + 3)}^{2} = 16$ $(x+3)^2=16$

How do you complete the square for x2+6x−7=0x^2+6x-7=0?