Answer
The solutions are $ \frac{7}{2} ± \sqrt\frac{37}{4}$, and the solution set is {${\frac{7}{2}+\sqrt\frac{37}{4}, \frac{7}{2}-\sqrt\frac{37}{4}}$} or {$\frac{7}{2} ± \sqrt\frac{37}{4}$}.
Work Step by Step
$x^2 - 7x + 3 = 0$
Subtract $3$ from both sides to isolate the binomial $x^2 - 7x$.
$x^2 - 7x + 3 -3= 0-3$
$x^2 - 7x = -3$
The coefficient of the x-term is $-7$.
Half of $-7$ is $-\frac{7}{2}$, and $(-\frac{7}{2})^2 = \frac{49}{4}$.
Thus, add $\frac{49}{4}$ to both sides of the equation to complete the square.
$x^2 - 7x + \frac{49}{4}= -3 +\frac{49}{4}$
$x^2 - 7x + \frac{49}{4} = \frac{37}{4}$
$(x-\frac{7}{2})^2 = \frac{37}{4}$
$ x -\frac{7}{2} =\sqrt\frac{37}{4}$ or $ x -\frac{7}{2} =-\sqrt\frac{37}{4}$
$x = \frac{7}{2}+\sqrt\frac{37}{4}$ or $x = \frac{7}{2}-\sqrt\frac{37}{4}$
The solutions are $ \frac{7}{2} ± \sqrt\frac{37}{4}$, and the solution set is {${\frac{7}{2}+\sqrt\frac{37}{4}, \frac{7}{2}-\sqrt\frac{37}{4}}$} or {$\frac{7}{2} ± \sqrt\frac{37}{4}$}.
