## College Algebra (6th Edition)

The solutions are $x=\dfrac{7}{2}\pm\dfrac{\sqrt{61}}{2}$
$x(x-7)=3$ Evaluate the product on the left side of the equation: $x^{2}-7x=3$ Take $3$ to the left side: $x^{2}-7x-3=0$ Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ In this case, $a=1$, $b=-7$ and $c=-3$ Substitute the known values into the formula and evaluate: $x=\dfrac{-(-7)\pm\sqrt{(-7)^{2}-4(1)(-3)}}{2(1)}=\dfrac{7\pm\sqrt{49+12}}{2}=...$ $...=\dfrac{7\pm\sqrt{61}}{2}=\dfrac{7}{2}\pm\dfrac{\sqrt{61}}{2}$