## Intermediate Algebra (6th Edition)

The solutions are: $\\x_1\approx 2.4$ $\\x_2 \approx 0.6$
RECALL: A quadratic equation of the form $ax^2+bx+c=0$ can be solved using the quadratic formula $x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$ The quadratic equation $2x^2-6x+3=0$has: $a=2 \\b=-6 \\c=3$ Substitute these values into the quadratic formula to obtain: $x=\dfrac{-(-6) \pm \sqrt{(-6)^2-4(2)(3)}}{2(2)} \\x=\dfrac{6 \pm \sqrt{36-24}}{4} \\x=\dfrac{6\pm \sqrt{12}}{4} \\x =\dfrac{6}{4} \pm \dfrac{\sqrt{12}}{4} \\x=\dfrac{3}{2} \pm \dfrac{\sqrt{12}}{4}$ Thus, $\\x_1=\dfrac{3}{2}+\dfrac{\sqrt{12}}{4} \approx 2.4$ $\\x_2=\dfrac{3}{2}-\dfrac{\sqrt{12}}{4} \approx 0.6$