3.4 Notes Part 1_Alg2

Put equation in standard form:  $ax^2+bx+c=0$  

 $2x^2-x-6=0$  

Find a, b, and c:   $a=2,\ b=-1,\ c=-6$  

Plug a, b, and c into the quadratic formula and solve

 $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$  

 $x=\frac{-\left(-1\right)\pm\sqrt{\left(-1\right)^2-4\left(2\right)\left(-6\right)}}{2\left(2\right)}$  

 $x=\frac{1\pm\sqrt{1+48}}{4}=\frac{1\pm\sqrt{49}}{4}$  

 $x\ =\ \frac{1\pm7}{4}$  

 $x=\frac{1+7}{4}=\frac{8}{4}=2\ \ \ or\ x=\frac{1-7}{4}=-\frac{6}{4}=-\frac{3}{2}$  

 $x=2\ or\ x=-\frac{3}{2}$  

 $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$  

 $r=\frac{-\left(3\right)\pm\sqrt{\left(3\right)^2-4\left(1\right)\left(-3\right)}}{2\left(1\right)}$  

 $r=\frac{-3\pm\sqrt{9+12}}{2}$  

 $r=\frac{-3\pm\sqrt{21}}{4}$  

 $r=\frac{-3+\sqrt{21}}{4}\ or\ \ r=\frac{-3-\sqrt{21}}{4}$