How do you solve $3 y^{2} - 5 x = 2 x - 4$ $3y ^ { 2} - 5x = 2x - 4$ ?

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Answer 1 · 2018-05-15T11:19:20

$y = \pm \sqrt{\frac{7 x - 4}{3}}$ $y=+-sqrt((7x-4)/3)$

$3 y^{2} - 5 x = 2 x - 4$ $3y^2-5x=2x-4$

$3 y^{2} - 5 x + 5 x = 2 x - 4 + 5 x = 7 x - 4$ $3y^2-5x+5x=2x-4+5x=7x-4$

$3 y^{2} = 7 x - 4$ $3y^2=7x-4$

$y^{2} = \frac{7 x - 4}{3}$ $y^2=(7x-4)/3$

$y = \pm \sqrt{\frac{7 x - 4}{3}}$ $y=+-sqrt((7x-4)/3)$

How do you solve 3y2−5x=2x−43y ^ { 2} - 5x = 2x - 4?