How do you solve by completing the square for $- 3 x^{2} + 8 x + 7 = 0$ $-3x^2 + 8x + 7 = 0$ ?

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Answer 1 · 2015-06-10T01:54:10

Solve: $y = - 3 x^{2} + 8 x + 7 = 0$ $y = -3x^2 + 8x + 7 = 0$

$y = - 3 (x^{2} - \frac{8 x}{3} + \frac{16}{9} - \frac{16}{9} - \frac{7}{3})$ $y = -3(x^2 - (8x)/3 + 16/9 - 16/9 - 7/3)$

$y = - 3 {(x - \frac{4}{3})}^{2} + \frac{16}{3} + 7$ $y = -3(x - 4/3)^2 + 16/3 + 7$

$- 3 {(x - \frac{4}{3})}^{2} = - \frac{16}{3} - 7 = - \frac{37}{3}$ $-3(x - 4/3)^2 = -16/3 - 7 = -37/3$

${(x - \frac{4}{3})}^{2} = \frac{37}{9} \to (x - \frac{4}{3}) = \pm \frac{\sqrt{37}}{3}$ $(x - 4/3)^2 = 37/9 -> (x - 4/3) = +- (sqrt37)/3$

$x = \frac{4}{3} \pm \frac{\sqrt{37}}{3}$ $x = 4/3 +- (sqrt37)/3$

How do you solve by completing the square for −3x2+8x+7=0-3x^2 + 8x + 7 = 0?