Solve x^2-3x/4=-1/4+6x^2-1/4

$$\frac { x ^ { 2 } - 3 x } { 4 } = \frac { - 1 } { 4 } + \frac { 6 x ^ { 2 } - 1 } { 4 }$$

Solve for x

$$x=-1$$

$$x=\frac{2}{5}=0.4$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 4.

$$x^{2}-3x=-1+6x^{2}-1$$

Subtract 1 from -1 to get -2.

$$x^{2}-3x=-2+6x^{2}$$

Subtract -2 from both sides.

$$x^{2}-3x-\left(-2\right)=6x^{2}$$

The opposite of -2 is 2.

$$x^{2}-3x+2=6x^{2}$$

Subtract 6x^{2} from both sides.

$$x^{2}-3x+2-6x^{2}=0$$

Combine x^{2} and -6x^{2} to get -5x^{2}.

$$-5x^{2}-3x+2=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -5x^{2}+ax+bx+2. To find a and b, set up a system to be solved.

$$a+b=-3 ab=-5\times 2=-10$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -10.

$$1,-10 2,-5$$

Calculate the sum for each pair.

$$1-10=-9 2-5=-3$$

The solution is the pair that gives sum -3.

$$a=2 b=-5$$

Rewrite -5x^{2}-3x+2 as \left(-5x^{2}+2x\right)+\left(-5x+2\right).

$$\left(-5x^{2}+2x\right)+\left(-5x+2\right)$$

Factor out -x in the first and -1 in the second group.

$$-x\left(5x-2\right)-\left(5x-2\right)$$

Factor out common term 5x-2 by using distributive property.

$$\left(5x-2\right)\left(-x-1\right)$$

To find equation solutions, solve 5x-2=0 and -x-1=0.

$$x=\frac{2}{5} x=-1$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 4.

$$x^{2}-3x=-1+6x^{2}-1$$

Subtract 1 from -1 to get -2.

$$x^{2}-3x=-2+6x^{2}$$

Subtract -2 from both sides.

$$x^{2}-3x-\left(-2\right)=6x^{2}$$

The opposite of -2 is 2.

$$x^{2}-3x+2=6x^{2}$$

Subtract 6x^{2} from both sides.

$$x^{2}-3x+2-6x^{2}=0$$

Combine x^{2} and -6x^{2} to get -5x^{2}.

$$-5x^{2}-3x+2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, -3 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-5\right)\times 2}}{2\left(-5\right)}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-5\right)\times 2}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-\left(-3\right)±\sqrt{9+20\times 2}}{2\left(-5\right)}$$

Multiply 20 times 2.

$$x=\frac{-\left(-3\right)±\sqrt{9+40}}{2\left(-5\right)}$$

Add 9 to 40.

$$x=\frac{-\left(-3\right)±\sqrt{49}}{2\left(-5\right)}$$

Take the square root of 49.

$$x=\frac{-\left(-3\right)±7}{2\left(-5\right)}$$

The opposite of -3 is 3.

$$x=\frac{3±7}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{3±7}{-10}$$

Now solve the equation x=\frac{3±7}{-10} when ± is plus. Add 3 to 7.

$$x=\frac{10}{-10}$$

Divide 10 by -10.

$$x=-1$$

Now solve the equation x=\frac{3±7}{-10} when ± is minus. Subtract 7 from 3.

$$x=-\frac{4}{-10}$$

Reduce the fraction \frac{-4}{-10} to lowest terms by extracting and canceling out 2.

$$x=\frac{2}{5}$$

The equation is now solved.

$$x=-1 x=\frac{2}{5}$$

Steps for Completing the Square

Multiply both sides of the equation by 4.

$$x^{2}-3x=-1+6x^{2}-1$$

Subtract 1 from -1 to get -2.

$$x^{2}-3x=-2+6x^{2}$$

Subtract 6x^{2} from both sides.

$$x^{2}-3x-6x^{2}=-2$$

Combine x^{2} and -6x^{2} to get -5x^{2}.

$$-5x^{2}-3x=-2$$

Divide both sides by -5.

$$\frac{-5x^{2}-3x}{-5}=-\frac{2}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\left(-\frac{3}{-5}\right)x=-\frac{2}{-5}$$

Divide -3 by -5.

$$x^{2}+\frac{3}{5}x=-\frac{2}{-5}$$

Divide -2 by -5.

$$x^{2}+\frac{3}{5}x=\frac{2}{5}$$

Divide \frac{3}{5}, the coefficient of the x term, by 2 to get \frac{3}{10}. Then add the square of \frac{3}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{3}{5}x+\left(\frac{3}{10}\right)^{2}=\frac{2}{5}+\left(\frac{3}{10}\right)^{2}$$

Square \frac{3}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{3}{5}x+\frac{9}{100}=\frac{2}{5}+\frac{9}{100}$$

Add \frac{2}{5} to \frac{9}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{3}{5}x+\frac{9}{100}=\frac{49}{100}$$

Factor x^{2}+\frac{3}{5}x+\frac{9}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{10}\right)^{2}=\frac{49}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{10}\right)^{2}}=\sqrt{\frac{49}{100}}$$

Simplify.

$$x+\frac{3}{10}=\frac{7}{10} x+\frac{3}{10}=-\frac{7}{10}$$

Subtract \frac{3}{10} from both sides of the equation.

$$x=\frac{2}{5} x=-1$$