Answer:
C. (-4, 11) and (2, -1)
Step-by-step explanation:
[tex]x^2-5=-2x+3\\\\x^2+2x-8=0\\\\a=1\,,\ b=2\,,\ c=-8\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x=\dfrac{-2\pm\sqrt{2^2-4\cdot1\cdot(-8)}}{2\cdot1}=\dfrac{-2\pm\sqrt{4+32}}{2}=\dfrac{-2\pm\sqrt{36}}{2}=\dfrac{-2\pm6}{2}\\\\x_1=\dfrac{-2-6}2=\dfrac{-8}2=-4\\\\y_1=-2x_1+3=-2(-4)+3=8+3=11\\\\(-4\,,\ 11)\\\\x_2=\dfrac{-2+6}{2}=\dfrac{4}2=2\\\\y_2=-2x_2+3=-2\cdot2+3=-4+3=-1\\\\(2\,,\ -1)[/tex]
