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

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Answer 1 · 2016-08-26T02:54:25

$p=+-isqrt(2/3)$ .

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

Divide both sides by $3$ to get $p^2=-2/3$ .

Square root both sides to get $p=+-sqrt(-2/3)$ , which is equivalent to $p=+-isqrt(2/3)$ .

How do you solve the quadratic 3p^2-4=-63p^2-4=-6 using any method?