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To solve an equation of the form \(x^2 + bx + c = 0\), consider the expression \(\left(x + \frac{b}{2}\right)^2 + c\).

\(\left(x + \frac{b}{2}\right)^2\) expands to \(x^2 + \frac{b}{2}x + \frac{b}{2}x + \left(\frac{b}{2}\right)^2\)

\(= x^2 + bx + \left(\frac{b}{2}\right)^2\)

\(x^2 + bx + c = 0\) is the same as \(x^2 + bx + (\frac{b}{2})^2 - (\frac{b}{2})^2 + c = 0\)

\( = \left(x + \frac{b}{2}\right)^2 - (\frac{b}{2})^2 + c = 0\)

This can be rearranged to give \(\left(x + \frac{b}{2}\right)^2 = \left(\frac{b}{2}\right)^2 - c\) which can then be solved by taking the square root of both sides.

Solve \(x^2 + 6x - 10 = 0\) by completing the square.

As the coefficient of \(x\) is 6, to complete the square we use \((x + \frac{6}{2})^2 = (x + 3)^2\)

Expanding \((x + 3)^2\) gives

\((x + 3)^2 = x^2 + 6x + 9\)

So \((x + 3)^2 - 9 = x^2 + 6x\)

The equation we are solving is \(x^2 + 6x – 10 = 0\)

(By replacing \(x^2 + 6x\)), this can be rewritten as \((x + 3)^2 – 9 – 10 = 0\)

This simplifies to \((x + 3)^2 – 19 = 0\)

Rearrange this quadratic to get \((x + 3)^2\) alone on the left-hand side by adding 19 to both sides

\((x + 3)^2 = 19\)

\(x + 3 = \pm \sqrt{19}\)

\(x = -3 \pm \sqrt{19}\)

\(x = -3 \pm \sqrt{19}\)

The first solution is: \(x = -3 + \sqrt{19} = 1.36\) (3 sf)

The second solution is: \(x = -3 - \sqrt{19} = -7.36\) (3 sf)