Answer:
[tex]x\in\left\{4-2\sqrt{10}\ ,\ \,4+2\sqrt{10}\,\right\}[/tex]
Step-by-step explanation:
[tex]x^2+13=8x+37\\\\x^2-8x-24=0\\\\a=1\,,\ \ b=-8\,,\ \ c=-24\\\\x_1=\dfrac{-(-8)-\sqrt{(-8)^2-4\cdot1\cdot(-24)}}{2\cdot1}=\dfrac{8-\sqrt{64+96}}2=\dfrac{8-\sqrt{160}}2 =\\\\=\dfrac{8-4\sqrt{10}}2=\dfrac{2(4-2\sqrt{10})}2=4-2\sqrt{10} \\\\x_2=\dfrac{-(-8)-\sqrt{(-8)^2-4\cdot1\cdot(-24)}}{2\cdot1}=\dfrac{8+4\sqrt{10}}2=4+2\sqrt{10}[/tex]
