How do you solve $x^{2} + 9 x + 12 = 5 x + 6$ $x^{2}+9x+12=5x+6$ ?

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How do you solve $x^{2} + 9 x + 12 = 5 x + 6$ $x^{2}+9x+12=5x+6$ ?

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Answer 1 · 2017-09-29T20:05:57

$x = - \sqrt{2} (i + \sqrt{2}) and \sqrt{2} (i - \sqrt{2}))$ $x=-sqrt2(i+sqrt2) and sqrt2(i-sqrt2))$

Explanation:

$x^{2} + 9 x - 5 x + 12 - 6 = 0$ $x^2+9x-5x+12-6=0$
$x^{2} + 4 x + 6 = 0$ $x^2+4x+6=0$
$x = \frac{- b \pm \sqrt{b^{2} - (4 a c)}}{2 a}$ $x=(-b+-sqrt(b^2-(4ac)))/(2a)$
$x = \frac{- 4 \pm \sqrt{16 - (4 \cdot 6)}}{2}$ $x=(-4+-sqrt(16-(4*6)))/2$
$x = - 2 \pm \sqrt{-} 2$ $x=-2+-sqrt-2$
$= (- 2 - i \sqrt{2}) and (- 2 + i \sqrt{2})$ $=(-2-isqrt2) and (-2+isqrt2)$

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How do you solve $x^{2} + 9 x + 12 = 5 x + 6$ $x^{2}+9x+12=5x+6$ ?

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How do you solve $x^{2} + 9 x + 12 = 5 x + 6$ $x^{2}+9x+12=5x+6$ ?