Question

How do you solve `6x ^ { 2} - 4x = 5`?

Answer 1

`x=(2+sqrt(34))/6,` `(2-sqrt(34))/6`

Explanation:

Solve:

`6x^2-4x=5`

Subtract `5` from both sides.

`6x^2-4x-5=0`

This is a quadratic equation in standard form: `ax^2+bx+c`, where `a=6`, `b=-4`, and `c=-5`.

Use the quadratic formula to solve for `x`.

`x=(-b+-sqrt(b^2-4ac))/(2a)`

Insert the known values into the formula. Be mindful of the negatives.

`x=(-(-4)+-sqrt((-4)^2-4*6*(-5)))/(2*6)`

Simplify. Again, be mindful of the negatives.

`x=(4+-sqrt(16+120))/12`

`x=(4+-sqrt(136))/12`

Prime factorize `136`.

`x=(4+-sqrt(2xx2xx2xx17))/12`

Simplify the square root.

`x=(4+-sqrt(2^(2)xx2xx17))/12`

`x=(4+-2sqrt(34))/12`

Simplify the equation. Divide the numerator and denominator by `2`.

`x=(2+-sqrt(34))/6`

Solutions for `x`

`x=(2+sqrt(34))/6`

`x=(2-sqrt(34))/6`