# Chapter 1 - Equations and Inequalities - 1.4 Quadratic Equations - 1.4 Exercises - Page 121: 48

The solutions are $x=\dfrac{3}{2}\pm\dfrac{\sqrt{3}}{6}i$

#### Work Step by Step

$-3x^{2}+9x=7$ Divide the whole equation by $-3$: $-\dfrac{1}{3}(-3x^{2}+9x=7)$ $x^{2}-3x=-\dfrac{7}{3}$ Complete the square by adding $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation and simplifying. In this case, $b=-3$ $x^{2}-3x+\Big(-\dfrac{3}{2}\Big)^{2}=-\dfrac{7}{3}+\Big(-\dfrac{3}{2}\Big)^{2}$ $x^{2}-3x+\dfrac{9}{4}=-\dfrac{7}{3}+\dfrac{9}{4}$ $x^{2}-3x+\dfrac{9}{4}=-\dfrac{1}{12}$ Factor the left side of the equation, which is a perfect square trinomial: $\Big(x-\dfrac{3}{2}\Big)^{2}=-\dfrac{1}{12}$ Take the square root of both sides: $\sqrt{\Big(x-\dfrac{3}{2}\Big)^{2}}=\pm\sqrt{-\dfrac{1}{12}}$ $x-\dfrac{3}{2}=\pm\dfrac{1}{2\sqrt{3}}i$ $x-\dfrac{3}{2}=\pm\dfrac{\sqrt{3}}{6}i$ Solve for $x$: $x=\dfrac{3}{2}\pm\dfrac{\sqrt{3}}{6}i$

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