Question

Solve the following equation, using the Bhaskara formula: x² - 5x + 6 = 0

Answer 1

    $x^2-5x+6=0$

    $x=\frac{-(-5)\pm\sqrt{(-5)^2-4\times1\times6}}{2\times1}\Leftrightarrow$

$\Leftrightarrow x=\frac{5\pm\sqrt{25-4\times6}}{2}\Leftrightarrow$

$\Leftrightarrow x=\frac{5\pm\sqrt{25-24}}{2}\Leftrightarrow$

$\Leftrightarrow x=\frac{5\pm\sqrt{1}}{2}\Leftrightarrow$

$\Leftrightarrow x=\frac{5\pm1}{2}\Leftrightarrow$

$\Leftrightarrow x=\frac{5-1}{2}\;\;\;\vee\;\;\;x=\frac{5+1}{2}\Leftrightarrow$

$\Leftrightarrow x=\frac{4}{2}\;\;\;\vee\;\;\;x=\frac{6}{2}\Leftrightarrow$

$\Leftrightarrow x=2\;\;\;\vee\;\;\;x=3$

Answer:  $x\in\{2\;;\;3\}$

Answer 2

$\sf\dfrac{-(-5)\pm\sqrt{\sf (-5)^2-4\cdot1\cdot6}}{2\cdot1}$

$\sf\dfrac{5\pm\sqrt{\sf 25-24}}{2}$

$\sf\dfrac{5\pm\sqrt{\sf 1}}{2}$

$\sf\dfrac{5\pm1}{2}$

$\sf x'=\dfrac{5-1}{2}~~~~~~x''=\dfrac{5+1}{2}$

$\sf x'=\dfrac{4}{2}~~~~~~~~~~~~x''=\dfrac{6}{2}$

$\sf x'=2~~~~~~~~~~~~~x''=3$

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Solutions = {2, 3}

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I hope that helps !!