Question

In the following exercises, solve by completing the square.

$3x^2-14x+8=0$

Short answer

The solution is $x=\frac{2}{3},x=4$

Step-by-step solution

Step 1. Given information 

Equation:$3x^2-14x+8=0$

Step 2. Calculation    

We have,

$3x^2-14x+8=0$

localid="1645178760253" $3x^2-14x+8-8=0-8$

$\frac{3x^2-14x}{3}=\frac{-8}{3}$

$x^2-\frac{14x}{3}=\frac{-8}{3}$

localid="1645178864584" $x^2-\frac{14x}{3}+{\left(\frac{1}{2}·\left(-\frac{14}{3}\right)\right)}^2=\frac{-8}{3}+{\left(\frac{1}{2}·\left(-\frac{14}{3}\right)\right)}^2$

localid="1645178889066" $x^2-\frac{14x}{3}+{\left(\frac{-7}{3}\right)}^2=\frac{-8}{3}+{\left(\frac{-7}{3}\right)}^2$

localid="1645179042680" $x^2-\frac{14x}{3}+\frac{49}{9}=\frac{-8}{3}+\frac{49}{9}$

localid="1645179082190" ${\left(x-\frac{7}{3}\right)}^2=\frac{25}{9}$

$\sqrt{{\left(x-\frac{7}{3}\right)}^2}=\sqrt{\frac{25}{9}}$

localid="1645179299091" $x-\frac{7}{3}=\pm \frac{5}{3}$

$x-\frac{7}{3}+\frac{7}{3}=-\frac{5}{3}+\frac{7}{3},x-\frac{7}{3}+\frac{7}{3}=\frac{5}{3}+\frac{7}{3}$

localid="1645179204367" $x=\frac{2}{3},x=4$