Answer:
[tex] x = \dfrac{ 3\pm \sqrt7}{2} [/tex]
Step-by-step explanation:
The given equation to us ,
[tex]\implies 6x = 2x^2 + 1 [/tex]
So we will use quadratic formula and solve out for x .
[tex]\implies 6x = 2x^2 + 1 \\\\\implies 2x^2 -6x + 1 = 0 \\\\\implies x = \dfrac{-b\pm \sqrt{ b^2-4ac}}{2a} \\\\\implies x = \dfrac{ 6 \pm \sqrt{ (-6)^2-4(2)(1) }}{2(2)} \\\\\implies x = \dfrac{ 6 \pm \sqrt{ 36 - 8 }}{4} \\\\\implies x = \dfrac{ 6 \pm \sqrt{ 28}}{4} \\\\\implies \boxed{\boxed{ x = \dfrac{ 3\pm \sqrt7}{2} }}[/tex]
