## Precalculus (6th Edition) Blitzer

The solution set of the equation $5{{x}^{2}}-6x-8=0$ is $\left\{ -\frac{4}{5},2 \right\}$.
Consider the equation, $5{{x}^{2}}-6x-8=0$ Compare the equation $5{{x}^{2}}-6x-8=0$ with the equation $a{{x}^{2}}+bx+c=0$ to find the values of $a$, $b$ and $c$. \begin{align} & a=5 \\ & b=-6 \\ & c=-8 \\ \end{align} Substitute $a=5$, $b=-6$ and $c=-8$ in the equation $x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. \begin{align} & x=\frac{-\left( -6 \right)\pm \sqrt{{{\left( -6 \right)}^{2}}-4\left( 5 \right)\left( -8 \right)}}{2\left( 5 \right)} \\ & =\frac{6\pm \sqrt{36+160}}{10} \\ & =\frac{6\pm \sqrt{196}}{10} \\ & =\frac{6\pm 14}{10} \end{align} Further solve the equation. \begin{align} & x=\frac{6-14}{10}\text{ or }x=\frac{6+14}{10} \\ & x=-\frac{8}{10}\text{ or }x=\frac{20}{10} \\ & x=-\frac{4}{5}\text{ or }x=2 \\ \end{align} Therefore, the solution set of the equation $5{{x}^{2}}-6x-8=0$ is $\left\{ -\frac{4}{5},2 \right\}$.