Resposta :
[tex]a)\;x+x^{2}=90\Longrightarrow\;x^{2}+x-90=0\\\\x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}\\\\x=\dfrac{-1\pm\sqrt{1^{2}-4\cdot1\cdot(-90)}}{2\cdot1}\\\\x=\dfrac{-1\pm\sqrt{361}}{2}\\\\x=\dfrac{-1\pm19}{2}\\\\x_{1}=-10\;\;\;\;\;\;\;x_{2}=9[/tex]
[tex]b)\;x^{2}+x=12\Longrightarrow\;x^{2}+x-12=0\\\\x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}\\\\x=\dfrac{-1\pm\sqrt{1^{2}-4\cdot1\cdot(-12)}}{2\cdot1}\\\\x=\dfrac{-1\pm\sqrt{49}}{2}\\\\x=\dfrac{-1\pm7}{2}\\\\x_{1}=3\;\;\;\;\;\;\;x_{2}=-4[/tex]
[tex]c)\;x^{2}-2x=-1\Longrightarrow\;x^{2}-2x+1=0\\\\x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}\\\\x=\dfrac{2\pm\sqrt{(-2)^{2}-4\cdot1\cdot1}}{2\cdot1}\\\\x=\dfrac{2\pm\sqrt{0}}{2}\\\\x_{1}=x_{2}=1[/tex]
[tex]d)[/tex] Não compreendi o que você escreveu
[tex]e)\;x^{2}+25=10x\Longrightarrow\;x^{2}-10x+25=0\\\\x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}\\\\x=\dfrac{10\pm\sqrt{(-10)^{2}-4\cdot1\cdot(25)}}{2\cdot1}\\\\x=\dfrac{10\pm\sqrt{0}}{2}\\\\x=\dfrac{10}{2}\\\\x_{1}=x_{2}=5[/tex]
[tex]b)\;x^{2}+x=12\Longrightarrow\;x^{2}+x-12=0\\\\x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}\\\\x=\dfrac{-1\pm\sqrt{1^{2}-4\cdot1\cdot(-12)}}{2\cdot1}\\\\x=\dfrac{-1\pm\sqrt{49}}{2}\\\\x=\dfrac{-1\pm7}{2}\\\\x_{1}=3\;\;\;\;\;\;\;x_{2}=-4[/tex]
[tex]c)\;x^{2}-2x=-1\Longrightarrow\;x^{2}-2x+1=0\\\\x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}\\\\x=\dfrac{2\pm\sqrt{(-2)^{2}-4\cdot1\cdot1}}{2\cdot1}\\\\x=\dfrac{2\pm\sqrt{0}}{2}\\\\x_{1}=x_{2}=1[/tex]
[tex]d)[/tex] Não compreendi o que você escreveu
[tex]e)\;x^{2}+25=10x\Longrightarrow\;x^{2}-10x+25=0\\\\x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}\\\\x=\dfrac{10\pm\sqrt{(-10)^{2}-4\cdot1\cdot(25)}}{2\cdot1}\\\\x=\dfrac{10\pm\sqrt{0}}{2}\\\\x=\dfrac{10}{2}\\\\x_{1}=x_{2}=5[/tex]