Quero que resolvam essa equação:
0,2n^2 + 1,8n - 126 = 0
Resolução completa !!!
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Olá.
![\mathsf{0,2n^2 + 1,8n - 126 = 0}\\\\\\
\mathsf{\Delta=b^2-4ac}\\\mathsf{\Delta=1,8^2-4\cdot[0,2\cdot(-126)]}\\\mathsf{\Delta=3,24-4\cdot[-25,2]}\\\mathsf{\Delta=3,24+100,8}\\\mathsf{\Delta=104,04}\\\\\mathsf{\sqrt{\Delta}=\sqrt{104,04}=10,2}\\\\\\\mathsf{x=\dfrac{-b\pm\sqrt{\Delta}}{2a}=\dfrac{-1,8\pm10,2}{0,4}}\\\\\\\mathsf{x'=\dfrac{-1,8+10,2}{0,4}=\dfrac{8,4}{0,4}=\boxed{21}}\\\\\\\mathsf{x''=\dfrac{-1,8-10,2}{0,4}=\dfrac{-12}{0,4}}=\boxed{-30}\\\\\\\mathsf{S=\{\-30, 21\}}](https://tex.z-dn.net/?f=%5Cmathsf%7B0%2C2n%5E2+%2B+1%2C8n+-+126+%3D+0%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B%5CDelta%3Db%5E2-4ac%7D%5C%5C%5Cmathsf%7B%5CDelta%3D1%2C8%5E2-4%5Ccdot%5B0%2C2%5Ccdot%28-126%29%5D%7D%5C%5C%5Cmathsf%7B%5CDelta%3D3%2C24-4%5Ccdot%5B-25%2C2%5D%7D%5C%5C%5Cmathsf%7B%5CDelta%3D3%2C24%2B100%2C8%7D%5C%5C%5Cmathsf%7B%5CDelta%3D104%2C04%7D%5C%5C%5C%5C%5Cmathsf%7B%5Csqrt%7B%5CDelta%7D%3D%5Csqrt%7B104%2C04%7D%3D10%2C2%7D%5C%5C%5C%5C%5C%5C%5Cmathsf%7Bx%3D%5Cdfrac%7B-b%5Cpm%5Csqrt%7B%5CDelta%7D%7D%7B2a%7D%3D%5Cdfrac%7B-1%2C8%5Cpm10%2C2%7D%7B0%2C4%7D%7D%5C%5C%5C%5C%5C%5C%5Cmathsf%7Bx%27%3D%5Cdfrac%7B-1%2C8%2B10%2C2%7D%7B0%2C4%7D%3D%5Cdfrac%7B8%2C4%7D%7B0%2C4%7D%3D%5Cboxed%7B21%7D%7D%5C%5C%5C%5C%5C%5C%5Cmathsf%7Bx%27%27%3D%5Cdfrac%7B-1%2C8-10%2C2%7D%7B0%2C4%7D%3D%5Cdfrac%7B-12%7D%7B0%2C4%7D%7D%3D%5Cboxed%7B-30%7D%5C%5C%5C%5C%5C%5C%5Cmathsf%7BS%3D%5C%7B%5C-30%2C+21%5C%7D%7D "\mathsf{0,2n^2 + 1,8n - 126 = 0}\\\\\\
\mathsf{\Delta=b^2-4ac}\\\mathsf{\Delta=1,8^2-4\cdot[0,2\cdot(-126)]}\\\mathsf{\Delta=3,24-4\cdot[-25,2]}\\\mathsf{\Delta=3,24+100,8}\\\mathsf{\Delta=104,04}\\\\\mathsf{\sqrt{\Delta}=\sqrt{104,04}=10,2}\\\\\\\mathsf{x=\dfrac{-b\pm\sqrt{\Delta}}{2a}=\dfrac{-1,8\pm10,2}{0,4}}\\\\\\\mathsf{x'=\dfrac{-1,8+10,2}{0,4}=\dfrac{8,4}{0,4}=\boxed{21}}\\\\\\\mathsf{x''=\dfrac{-1,8-10,2}{0,4}=\dfrac{-12}{0,4}}=\boxed{-30}\\\\\\\mathsf{S=\{\-30, 21\}}")
A imagem é pra caso o LaTex falhe.
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Bons estudos.
A imagem é pra caso o LaTex falhe.
Qualquer dúvida, deixe nos comentários.
Bons estudos.
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