Question

Como resolver x²-5x+6=0

Answer 1

$\large\boxed{\begin{array}{l}\rm x^2-5x+6=0\\\rm\Delta=b^2-4ac\\\rm\Delta=(-5)^2-4\cdot1\cdot6\\\rm\Delta=25-24\\\rm\Delta=1\\\rm x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\\\rm x=\dfrac{-(-5)\pm\sqrt{1}}{2\cdot1}\\\\\rm x=\dfrac{5\pm1}{2}\begin{cases}\rm x_1=\dfrac{5+1}{2}=\dfrac{6}{2}=3\\\\\rm x_2=\dfrac{ 5-1}{2}=\dfrac{4}{2}=2\end{cases}\\\\\rm S=\{2,3\}\end{array}}$

Answer 2

$\large\boxed{\begin{array}{l} \rm \: x {}^{2} - 5x + 6 = 0 \\ \\ \rm \: a = 1 \\ \rm \: b = - 5 \\ \rm \: c = 6 \\ \\ \rm\Delta = b {}^{2} - 4ac \\ \Delta = ( - 5) {}^{2} - 4.1.6 \\\Delta = 25 - 24 \\ \Delta = 1 \\ \\ \rm \: x = \dfrac{ - b \pm \sqrt{\Delta} }{2a} \\ \\ \rm \: x = \dfrac{ - ( - 5) \pm \sqrt{1} }{2.1} \\ \\ \rm \: x = \dfrac{5 \pm1}{2} \begin{cases} \rm \: x_1 = \dfrac{5 + 1}{2} = \dfrac{6}{2} = 3 \\ \\ \rm \: x_2 = \dfrac{5 - 1}{2} = \dfrac{4}{2} = 2 \end{cases} \\ \\ \rm \: S= \{2,3 \} \end{array}}$