How do you solve $x^{2} - 3 x = 28$ $x^2-3x=28$ by completing the square?

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Answer 1 · 2018-03-03T07:48:49

$x = 7, - 4$ $color(green)(x = 7, -4$

$x^{2} - 3 x = 28$ $x^2 - 3x = 28$

Add ${(\frac{3}{2})}^{2}$ $(3/2)^2$ to both sides.

$x^{2} - (2 \cdot (\frac{3}{2}) \cdot x) + {(\frac{3}{2})}^{2} = 28 + {(\frac{3}{2})}^{2}$ $x^2 - (2 * (3/2) * x) + (3/2)^2 = 28 + (3/2)^2$

${(x - \frac{3}{2})}^{2} = \frac{121}{4} = {(\frac{11}{2})}^{2}$ $(x-3/2)^2 = 121 /4 = (11/2)^2$

$x - \frac{3}{2} = \pm \frac{11}{2}$ $x - 3/2 = +- 11 /2$

$x = \pm \frac{11}{2} + \frac{3}{2}$ $x = +- 11/2 + 3/2$

$x = 7, - 4$ $color(green)(x = 7, -4$

How do you solve x^2-3x=28x2−3x=28x^2-3x=28 by completing the square?