Precalculus (6th Edition) Blitzer

Step 1. Rewrite the equation as $3y^2-(6x+8)y+(3x^2+10x-2)$ and let $a=3, b=-(6x+8), c=3x^2+10x-2$ Step 2. Use the quadratic formula $y=\frac{-b\pm\sqrt {b^2-4ac}}{2a}$ to get $y=\frac{6x+8\pm\sqrt {(6x+8)^2-4(3)(3x^2+10x-2)}}{6}=\frac{3x+4\pm\sqrt {22-6x}}{3}$ Step 3. We can graph the equations $y_1=\frac{3x+4+\sqrt {22-6x}}{3}$ and $y_2=\frac{3x+4-\sqrt {22-6x}}{3}$ as shown in the figure.