## Precalculus (6th Edition) Blitzer

The solution set of the equation is$\left\{ \frac{3}{2},-2 \right\}$
Consider the equation: $2{{x}^{2}}+x=6$ Re-write the above equation as, $2{{x}^{2}}+x-6=0$ Now factor: \begin{align} & 2{{x}^{2}}+4x-3x-6=0 \\ & 2x\left( x+2 \right)-3\left( x+2 \right)=0 \\ & \left( 2x-3 \right)\left( x+2 \right)=0 \end{align} Then equate each term to zero, \begin{align} & 2x-3=0 \\ & 2x=3 \\ & x=\frac{3}{2} \end{align} And, \begin{align} & x+2=0 \\ & x=-2 \end{align} Therefore, the value of $x=\frac{3}{2}\text{ and }x=-2$ Check: Put $\left( x=\frac{3}{2} \right)$ in the provided equation and verify, \begin{align} & 2{{\left( \frac{3}{2} \right)}^{2}}+\frac{3}{2}\overset{?}{\mathop{=}}\,6 \\ & 2\left( \frac{9}{4} \right)+\frac{3}{2}\overset{?}{\mathop{=}}\,6 \\ & \frac{9}{2}+\frac{3}{2}\overset{?}{\mathop{=}}\,6 \end{align} Further, simplify \begin{align} \frac{9}{2}+\frac{3}{2}\overset{?}{\mathop{=}}\,6 & \\ \frac{12}{2}\overset{?}{\mathop{=}}\,6 & \\ 6=6 & \\ \end{align} This is true. Check: Put $\left( x=-2 \right)$ in the provided equation and verify, \begin{align} & 2{{\left( -2 \right)}^{2}}+\left( -2 \right)\overset{?}{\mathop{=}}\,6 \\ & 2\left( 4 \right)-2\overset{?}{\mathop{=}}\,6 \\ & 8-2\overset{?}{\mathop{=}}\,6 \\ & 6=6 \end{align} This is also true. Hence, the solution of the equation is $\left\{ \frac{3}{2},-2 \right\}$.