Variable $x$ cannot be equal to $\frac{1}{2}$ since division by zero is not defined. Multiply both sides of the equation by $2x-1$.
$$6x^{2}+5x-4=2x-1$$
Subtract $2x$ from both sides.
$$6x^{2}+5x-4-2x=-1$$
Combine $5x$ and $-2x$ to get $3x$.
$$6x^{2}+3x-4=-1$$
Add $1$ to both sides.
$$6x^{2}+3x-4+1=0$$
Add $-4$ and $1$ to get $-3$.
$$6x^{2}+3x-3=0$$
Divide both sides by $3$.
$$2x^{2}+x-1=0$$
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $2x^{2}+ax+bx-1$. To find $a$ and $b$, set up a system to be solved.
$$a+b=1$$ $$ab=2\left(-1\right)=-2$$
Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.
$$a=-1$$ $$b=2$$
Rewrite $2x^{2}+x-1$ as $\left(2x^{2}-x\right)+\left(2x-1\right)$.
$$\left(2x^{2}-x\right)+\left(2x-1\right)$$
Factor out $x$ in $2x^{2}-x$.
$$x\left(2x-1\right)+2x-1$$
Factor out common term $2x-1$ by using distributive property.
$$\left(2x-1\right)\left(x+1\right)$$
To find equation solutions, solve $2x-1=0$ and $x+1=0$.
$$x=\frac{1}{2}$$ $$x=-1$$
Variable $x$ cannot be equal to $\frac{1}{2}$.
$$x=-1$$
Steps Using the Quadratic Formula
Variable $x$ cannot be equal to $\frac{1}{2}$ since division by zero is not defined. Multiply both sides of the equation by $2x-1$.
$$6x^{2}+5x-4=2x-1$$
Subtract $2x$ from both sides.
$$6x^{2}+5x-4-2x=-1$$
Combine $5x$ and $-2x$ to get $3x$.
$$6x^{2}+3x-4=-1$$
Add $1$ to both sides.
$$6x^{2}+3x-4+1=0$$
Add $-4$ and $1$ to get $-3$.
$$6x^{2}+3x-3=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $6$ for $a$, $3$ for $b$, and $-3$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
Now solve the equation $x=\frac{-3±9}{12}$ when $±$ is plus. Add $-3$ to $9$.
$$x=\frac{6}{12}$$
Reduce the fraction $\frac{6}{12}$ to lowest terms by extracting and canceling out $6$.
$$x=\frac{1}{2}$$
Now solve the equation $x=\frac{-3±9}{12}$ when $±$ is minus. Subtract $9$ from $-3$.
$$x=-\frac{12}{12}$$
Divide $-12$ by $12$.
$$x=-1$$
The equation is now solved.
$$x=\frac{1}{2}$$ $$x=-1$$
Variable $x$ cannot be equal to $\frac{1}{2}$.
$$x=-1$$
Steps for Completing the Square
Variable $x$ cannot be equal to $\frac{1}{2}$ since division by zero is not defined. Multiply both sides of the equation by $2x-1$.
$$6x^{2}+5x-4=2x-1$$
Subtract $2x$ from both sides.
$$6x^{2}+5x-4-2x=-1$$
Combine $5x$ and $-2x$ to get $3x$.
$$6x^{2}+3x-4=-1$$
Add $4$ to both sides.
$$6x^{2}+3x=-1+4$$
Add $-1$ and $4$ to get $3$.
$$6x^{2}+3x=3$$
Divide both sides by $6$.
$$\frac{6x^{2}+3x}{6}=\frac{3}{6}$$
Dividing by $6$ undoes the multiplication by $6$.
$$x^{2}+\frac{3}{6}x=\frac{3}{6}$$
Reduce the fraction $\frac{3}{6}$ to lowest terms by extracting and canceling out $3$.
$$x^{2}+\frac{1}{2}x=\frac{3}{6}$$
Reduce the fraction $\frac{3}{6}$ to lowest terms by extracting and canceling out $3$.
$$x^{2}+\frac{1}{2}x=\frac{1}{2}$$
Divide $\frac{1}{2}$, the coefficient of the $x$ term, by $2$ to get $\frac{1}{4}$. Then add the square of $\frac{1}{4}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.
Add $\frac{1}{2}$ to $\frac{1}{16}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{9}{16}$$
Factor $x^{2}+\frac{1}{2}x+\frac{1}{16}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.
$$\left(x+\frac{1}{4}\right)^{2}=\frac{9}{16}$$
Take the square root of both sides of the equation.
