Solve 12x^2-26x-7=3x+1

$$12 x ^ { 2 } - 26 x - 7 = 3 x + 1$$

Solve for x

$$x=-\frac{1}{4}=-0.25$$

$$x = \frac{8}{3} = 2\frac{2}{3} \approx 2.666666667$$

Steps Using Factoring By Grouping

Subtract 3x from both sides.

$$12x^{2}-26x-7-3x=1$$

Combine -26x and -3x to get -29x.

$$12x^{2}-29x-7=1$$

Subtract 1 from both sides.

$$12x^{2}-29x-7-1=0$$

Subtract 1 from -7 to get -8.

$$12x^{2}-29x-8=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 12x^{2}+ax+bx-8. To find a and b, set up a system to be solved.

$$a+b=-29 ab=12\left(-8\right)=-96$$

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 -96.

$$1,-96 2,-48 3,-32 4,-24 6,-16 8,-12$$

Calculate the sum for each pair.

$$1-96=-95 2-48=-46 3-32=-29 4-24=-20 6-16=-10 8-12=-4$$

The solution is the pair that gives sum -29.

$$a=-32 b=3$$

Rewrite 12x^{2}-29x-8 as \left(12x^{2}-32x\right)+\left(3x-8\right).

$$\left(12x^{2}-32x\right)+\left(3x-8\right)$$

Factor out 4x in 12x^{2}-32x.

$$4x\left(3x-8\right)+3x-8$$

Factor out common term 3x-8 by using distributive property.

$$\left(3x-8\right)\left(4x+1\right)$$

To find equation solutions, solve 3x-8=0 and 4x+1=0.

$$x=\frac{8}{3} x=-\frac{1}{4}$$

Steps Using the Quadratic Formula

Subtract 3x from both sides.

$$12x^{2}-26x-7-3x=1$$

Combine -26x and -3x to get -29x.

$$12x^{2}-29x-7=1$$

Subtract 1 from both sides.

$$12x^{2}-29x-7-1=0$$

Subtract 1 from -7 to get -8.

$$12x^{2}-29x-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 12 for a, -29 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-29\right)±\sqrt{\left(-29\right)^{2}-4\times 12\left(-8\right)}}{2\times 12}$$

Square -29.

$$x=\frac{-\left(-29\right)±\sqrt{841-4\times 12\left(-8\right)}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-\left(-29\right)±\sqrt{841-48\left(-8\right)}}{2\times 12}$$

Multiply -48 times -8.

$$x=\frac{-\left(-29\right)±\sqrt{841+384}}{2\times 12}$$

Add 841 to 384.

$$x=\frac{-\left(-29\right)±\sqrt{1225}}{2\times 12}$$

Take the square root of 1225.

$$x=\frac{-\left(-29\right)±35}{2\times 12}$$

The opposite of -29 is 29.

$$x=\frac{29±35}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{29±35}{24}$$

Now solve the equation x=\frac{29±35}{24} when ± is plus. Add 29 to 35.

$$x=\frac{64}{24}$$

Reduce the fraction \frac{64}{24} to lowest terms by extracting and canceling out 8.

$$x=\frac{8}{3}$$

Now solve the equation x=\frac{29±35}{24} when ± is minus. Subtract 35 from 29.

$$x=-\frac{6}{24}$$

Reduce the fraction \frac{-6}{24} to lowest terms by extracting and canceling out 6.

$$x=-\frac{1}{4}$$

The equation is now solved.

$$x=\frac{8}{3} x=-\frac{1}{4}$$

Steps for Completing the Square

Subtract 3x from both sides.

$$12x^{2}-26x-7-3x=1$$

Combine -26x and -3x to get -29x.

$$12x^{2}-29x-7=1$$

Add 7 to both sides.

$$12x^{2}-29x=1+7$$

Add 1 and 7 to get 8.

$$12x^{2}-29x=8$$

Divide both sides by 12.

$$\frac{12x^{2}-29x}{12}=\frac{8}{12}$$

Dividing by 12 undoes the multiplication by 12.

$$x^{2}-\frac{29}{12}x=\frac{8}{12}$$

Reduce the fraction \frac{8}{12} to lowest terms by extracting and canceling out 4.

$$x^{2}-\frac{29}{12}x=\frac{2}{3}$$

Divide -\frac{29}{12}, the coefficient of the x term, by 2 to get -\frac{29}{24}. Then add the square of -\frac{29}{24} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{29}{12}x+\left(-\frac{29}{24}\right)^{2}=\frac{2}{3}+\left(-\frac{29}{24}\right)^{2}$$

Square -\frac{29}{24} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{29}{12}x+\frac{841}{576}=\frac{2}{3}+\frac{841}{576}$$

Add \frac{2}{3} to \frac{841}{576} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{29}{12}x+\frac{841}{576}=\frac{1225}{576}$$

Factor x^{2}-\frac{29}{12}x+\frac{841}{576}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{29}{24}\right)^{2}=\frac{1225}{576}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{29}{24}\right)^{2}}=\sqrt{\frac{1225}{576}}$$

Simplify.

$$x-\frac{29}{24}=\frac{35}{24} x-\frac{29}{24}=-\frac{35}{24}$$

Add \frac{29}{24} to both sides of the equation.

$$x=\frac{8}{3} x=-\frac{1}{4}$$