## Algebra 1

The quadratic formula states that if a$x^{2}$+bx+c=0, and a$\ne$0, then $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. If $x^{2}$+8x+11=0, then a=1, b=8, and c=11, and to solve for x, then all we must do is plug the numbers into the quadratic formula then simplify. $x=\frac{-(8)\pm\sqrt{(8)^2-4(1)(11)}}{2(1)}$ =$\frac{-8\pm\sqrt{64-44)}}{2}$ (8 squared=64, 4*1*11=44, and 2*1=2) =$\frac{-8\pm\sqrt{20}}{2}$ (64-44=20) Since there is a plus minus symbol, we of course need to make sure that we compute both of our answers. Rounded to the nearest hundredth x=$\frac{-8+\sqrt 20}{2}$= -1.76 x=$\frac{-8-\sqrt 20}{2}$=-6.24