How do you solve ${(3 x - 2)}^{2} - 7 = 0$ $(3x-2)^2-7=0$ ?

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Answer 1 · 2015-10-03T21:03:13

This equation has 2 solutions:

$x_{1} = \frac{2 - \sqrt{7}}{3}$ $x_1=(2-sqrt(7))/3$

$x_{2} = \frac{2 + \sqrt{7}}{3}$ $x_2=(2+sqrt(7))/3$

${(3 x - 2)}^{2} - 7 = 0$ $(3x-2)^2-7=0$

$9 x^{2} - 12 x + 4 - 7 = 0$ $9x^2-12x+4-7=0$

$9 x^{2} - 12 x - 3 = 0$ $9x^2-12x-3=0$

$3 x^{2} - 4 x - 1 = 0$ $3x^2-4x-1=0$

$Delta=(-4)^2-4*3*(-1)$

$Delta=16+12$

$Delta=28$

$sqrt(Delta)=2sqrt(7)$

$x_1=(4-2sqrt(7))/6$

$x_1=(2-sqrt(7))/3$

$x_2=(2+sqrt(7))/3$

How do you solve (3x-2)^2-7=0(3x−2)2−7=0(3x-2)^2-7=0?