Question

In the following exercises, solve by completing the square.

$2c^2+c=6$

Short answer

The solution is$c=-2,c=\frac{3}{2}$.

Step-by-step solution

Step 1. Given information 

Equation:$2c^2+c=6$

Step 2. Calculation   

We have,

$2c^2+c=6$

$\frac{2c^2+c}{2}=\frac{6}{2}$

$c^2+\frac{c}{2}=3$

$c^2+\frac{c}{2}+{\left(\frac{1}{2}·\left(\frac{1}{2}\right)\right)}^2=3+{\left(\frac{1}{2}·\left(\frac{1}{2}\right)\right)}^2$

$c^2+\frac{c}{2}+{\left(\frac{1}{4}\right)}^2=3+{\left(\frac{1}{4}\right)}^2$

$c^2+\frac{c}{2}+\frac{1}{16}=3+\frac{1}{16}$

role="math" localid="1645176675401" ${\left(c+\frac{1}{4}\right)}^2=\frac{49}{16}$

$\sqrt{{\left(c+\frac{1}{4}\right)}^2}=\sqrt{\frac{49}{16}}$

$c+\frac{1}{4}=\pm \frac{7}{4}$

$c+\frac{1}{4}-\frac{1}{4}=-\frac{7}{4}-\frac{1}{4},c+\frac{1}{4}-\frac{1}{4}=\frac{7}{4}-\frac{1}{4}$

$c=-\frac{8}{4},c=\frac{6}{4}$

$c=-2,c=\frac{3}{2}$