## Precalculus (6th Edition) Blitzer

The solution of the equation is $\left\{ \underline{\frac{2+\sqrt{3}i}{2},\frac{2-\sqrt{3}i}{2}} \right\}$
Represent the provided equation in standard form as shown below: $4{{x}^{2}}-8x+7=0$ The roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ are given by the quadratic formula $x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ Where, \begin{align} & a=4 \\ & b=-8 \\ & c=7 \\ \end{align} Substitute the values of a, b, and c in the formula and obtain, \begin{align} & x=\frac{-\left( -8 \right)\pm \sqrt{{{\left( -8 \right)}^{2}}-4\left( 4 \right)\left( 7 \right)}}{2\left( 4 \right)} \\ & x=\frac{8\pm \sqrt{-48}}{8} \\ & x=\frac{8\pm 4\sqrt{3}i}{8} \\ & x=\frac{2\pm \sqrt{3}i}{2} \\ \end{align}. Therefore, $x=\frac{2+\sqrt{3}i}{2},\frac{2-\sqrt{3}i}{2}$. Hence, the solution of the equation is $x=\left\{ \underline{\frac{2+\sqrt{3}i}{2},\frac{2-\sqrt{3}i}{2}} \right\}$.