How do you solve the quadratic $3 p^{2} - 4 = - 6$ $3p^2-4=-6$ using any method?

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How do you solve the quadratic $3 p^{2} - 4 = - 6$ $3p^2-4=-6$ using any method?

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Answer 1 · 2016-08-26T02:54:25

$p = \pm i \sqrt{\frac{2}{3}}$ $p=+-isqrt(2/3)$ .

Explanation:

Add $4$ $4$ to both sides to get $3 p^{2} = - 2$ $3p^2=-2$ .

Divide both sides by $3$ $3$ to get $p^{2} = - \frac{2}{3}$ $p^2=-2/3$ .

Square root both sides to get $p = \pm \sqrt{- \frac{2}{3}}$ $p=+-sqrt(-2/3)$ , which is equivalent to $p = \pm i \sqrt{\frac{2}{3}}$ $p=+-isqrt(2/3)$ .

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How do you solve the quadratic $3 p^{2} - 4 = - 6$ $3p^2-4=-6$ using any method?

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