Question

6⋅x2 = 5⋅x + 1 solve using quadratic equation

Answer 1

\[(-b \pm \sqrt{b^2-4ac} ) / 2a\]\[6x^2 = 5x + 1\]
First, subtract (5x+1) from both sides so that the right side is equal to zero. You could subtract 6x^2 from both sides but I prefer the leading term to be positive.
\[6x^2 - 5x - 1 = 0\]
This is a form of the equation ax+bx+c = 0, so
a = 6
b = -5
c = -1
Plug this into the equation above: \[(-b \pm \sqrt{b^2-4ac} ) / 2a\]
And you get
\([-(-5) \pm \sqrt{(-5)^2 - 4(6)(1)})/2(6)\]
Which simplifies to\[(5 \pm \sqrt{25 - 24}) / 12\]
or\[(5 \pm \sqrt{1}) / 12\]
Which is
\[(5 \pm 1)/12\]
Because the square root of 1 is 1 (1*1=1)
Now, you have to take both the + and - possibilities of the equation:

So, in short, x = 1/2 and x = 1/3