Question

6 ( $x^{2}$ - 1 ) - X = 14 + 5 $x^{2}$

Answer 1

$6 (x^2 - 1) - x = 14 + 5x \\ \\ \\ 6x^2 - 6 - x = 14 + 5x \\ \\ 6x^2 - x - 6 - 14 - 5x^2 = 0 \\ \\ 6x^2 - 5x^2 -x - 6 - 14 = 0 \\ \\ \\ => x^2- x - 20 = 0$

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$x = \dfrac{-b \pm \sqrt{b^2 -4*a*c}}{2*a}$

a=1, b= −1, c=−20

Δ=b²2−4ac
Δ=(−1)²−4*1*−20
Δ=1+80
Δ=81

$x = \dfrac{-b \pm \sqrt{\triangle}}{2*a}$

$x = \dfrac{-(-1) \pm \sqrt{81}}{2*1} \\ \\ \\ x = \dfrac{1 \pm 9}{2} \\ \\ \\ x' = \dfrac{1 + 9}{2} \\ \\ \\ x' = \dfrac{10}{2} \\ \\ \\ x' = 5 \\ \\ \\ x'' = \dfrac{1 - 9}{2} \\ \\ \\ x'' = \dfrac{-8}{2} \\ \\ \\ x'' = -4$


S = {5,  -4 }