How do you solve $5 x^{2} + 12 r = - 4 x^{2} - 4$ $5x^2 + 12r = -4x^2 -4$ ?

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Answer 1 · 2015-06-08T13:44:22

$5 x^{2} + 12 r = - 4 x^{2} - 4$ $5x^2+12r = -4x^2-4$

$\Rightarrow 9 x^{2} = - 12 r - 4$ $rArr 9x^2 = -12r-4$

$\Rightarrow {(3 x)}^{2} = - 12 r - 4$ $rArr (3x)^2 = -12r -4$

$\Rightarrow 3 x = \pm \sqrt{- 12 r - 4}$ $rArr 3x = +-sqrt(-12r-4)$

$\Rightarrow x = \pm \frac{\sqrt{- 12 r - 4}}{3}$ $rArr x = +- sqrt(-12r-4)/3$

How do you solve 5x2+12r=−4x2−45x^2 + 12r = -4x^2 -4?