Respuesta :
[tex]\begin{gathered} y=-x^2-2x+8 \\ y=x^2-8x-12 \end{gathered}[/tex]
start by making both expressions equal
[tex]x^2-8x-12=-x^2-2x+8[/tex]bring all to one side
[tex]\begin{gathered} x^2+x^2-8x+2x-12-8=0 \\ 2x^2-6x-20=0 \end{gathered}[/tex]use the quadratic formula to solve the quadratic equation using a as 2, b as -6, and c as -20.
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]then,
[tex]\begin{gathered} x=\frac{-(-6)\pm\sqrt[]{(-6)^2-4\cdot(2)(-20)}}{2\cdot2} \\ x=\frac{6\pm\sqrt[]{36+160}}{4} \\ x=\frac{6\pm\sqrt[]{196}}{4} \\ x=\frac{6\pm14}{4} \\ \end{gathered}[/tex]since the expression is a quadratic equation there are going to be two possible solutions
[tex]\begin{gathered} x_1=\frac{6+14}{4} \\ x_1=\frac{20}{4}=5 \\ x_2=\frac{6-14}{4} \\ x_2=\frac{-8}{4}=-2 \end{gathered}[/tex]then, replace both -2 and 5 into one of the equations
[tex]\begin{gathered} \text{for x=-2} \\ y=-(-2)^2-2(-2)+8 \\ y=-4+4+8 \\ y=8 \\ \text{for x=5} \\ y=5^2-8\cdot5-12 \\ y=25-40-12 \\ y=-27 \end{gathered}[/tex]the solutions to the system are (-2,8) and (5,-27)
