Solve (4x+1)(4x+1)-(4x+1)(7x-6)=0

$$( 4 x + 1 ) ( 4 x + 1 ) - ( 4 x + 1 ) ( 7 x - 6 ) = 0$$

Solve for x

$$x = \frac{7}{3} = 2\frac{1}{3} \approx 2.333333333$$

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

Steps Using Factoring By Grouping

Multiply 4x+1 and 4x+1 to get \left(4x+1\right)^{2}.

$$\left(4x+1\right)^{2}-\left(4x+1\right)\left(7x-6\right)=0$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4x+1\right)^{2}.

$$16x^{2}+8x+1-\left(4x+1\right)\left(7x-6\right)=0$$

Use the distributive property to multiply 4x+1 by 7x-6 and combine like terms.

$$16x^{2}+8x+1-\left(28x^{2}-17x-6\right)=0$$

To find the opposite of 28x^{2}-17x-6, find the opposite of each term.

$$16x^{2}+8x+1-28x^{2}+17x+6=0$$

Combine 16x^{2} and -28x^{2} to get -12x^{2}.

$$-12x^{2}+8x+1+17x+6=0$$

Combine 8x and 17x to get 25x.

$$-12x^{2}+25x+1+6=0$$

Add 1 and 6 to get 7.

$$-12x^{2}+25x+7=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -12x^{2}+ax+bx+7. To find a and b, set up a system to be solved.

$$a+b=25 ab=-12\times 7=-84$$

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. List all such integer pairs that give product -84.

$$-1,84 -2,42 -3,28 -4,21 -6,14 -7,12$$

Calculate the sum for each pair.

$$-1+84=83 -2+42=40 -3+28=25 -4+21=17 -6+14=8 -7+12=5$$

The solution is the pair that gives sum 25.

$$a=28 b=-3$$

Rewrite -12x^{2}+25x+7 as \left(-12x^{2}+28x\right)+\left(-3x+7\right).

$$\left(-12x^{2}+28x\right)+\left(-3x+7\right)$$

Factor out -4x in the first and -1 in the second group.

$$-4x\left(3x-7\right)-\left(3x-7\right)$$

Factor out common term 3x-7 by using distributive property.

$$\left(3x-7\right)\left(-4x-1\right)$$

To find equation solutions, solve 3x-7=0 and -4x-1=0.

$$x=\frac{7}{3} x=-\frac{1}{4}$$

Steps Using the Quadratic Formula

Multiply 4x+1 and 4x+1 to get \left(4x+1\right)^{2}.

$$\left(4x+1\right)^{2}-\left(4x+1\right)\left(7x-6\right)=0$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4x+1\right)^{2}.

$$16x^{2}+8x+1-\left(4x+1\right)\left(7x-6\right)=0$$

Use the distributive property to multiply 4x+1 by 7x-6 and combine like terms.

$$16x^{2}+8x+1-\left(28x^{2}-17x-6\right)=0$$

To find the opposite of 28x^{2}-17x-6, find the opposite of each term.

$$16x^{2}+8x+1-28x^{2}+17x+6=0$$

Combine 16x^{2} and -28x^{2} to get -12x^{2}.

$$-12x^{2}+8x+1+17x+6=0$$

Combine 8x and 17x to get 25x.

$$-12x^{2}+25x+1+6=0$$

Add 1 and 6 to get 7.

$$-12x^{2}+25x+7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -12 for a, 25 for b, and 7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-25±\sqrt{25^{2}-4\left(-12\right)\times 7}}{2\left(-12\right)}$$

Square 25.

$$x=\frac{-25±\sqrt{625-4\left(-12\right)\times 7}}{2\left(-12\right)}$$

Multiply -4 times -12.

$$x=\frac{-25±\sqrt{625+48\times 7}}{2\left(-12\right)}$$

Multiply 48 times 7.

$$x=\frac{-25±\sqrt{625+336}}{2\left(-12\right)}$$

Add 625 to 336.

$$x=\frac{-25±\sqrt{961}}{2\left(-12\right)}$$

Take the square root of 961.

$$x=\frac{-25±31}{2\left(-12\right)}$$

Multiply 2 times -12.

$$x=\frac{-25±31}{-24}$$

Now solve the equation x=\frac{-25±31}{-24} when ± is plus. Add -25 to 31.

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

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

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

Now solve the equation x=\frac{-25±31}{-24} when ± is minus. Subtract 31 from -25.

$$x=-\frac{56}{-24}$$

Reduce the fraction \frac{-56}{-24} to lowest terms by extracting and canceling out 8.

$$x=\frac{7}{3}$$

The equation is now solved.

$$x=-\frac{1}{4} x=\frac{7}{3}$$

Steps for Completing the Square

Multiply 4x+1 and 4x+1 to get \left(4x+1\right)^{2}.

$$\left(4x+1\right)^{2}-\left(4x+1\right)\left(7x-6\right)=0$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4x+1\right)^{2}.

$$16x^{2}+8x+1-\left(4x+1\right)\left(7x-6\right)=0$$

Use the distributive property to multiply 4x+1 by 7x-6 and combine like terms.

$$16x^{2}+8x+1-\left(28x^{2}-17x-6\right)=0$$

To find the opposite of 28x^{2}-17x-6, find the opposite of each term.

$$16x^{2}+8x+1-28x^{2}+17x+6=0$$

Combine 16x^{2} and -28x^{2} to get -12x^{2}.

$$-12x^{2}+8x+1+17x+6=0$$

Combine 8x and 17x to get 25x.

$$-12x^{2}+25x+1+6=0$$

Add 1 and 6 to get 7.

$$-12x^{2}+25x+7=0$$

Subtract 7 from both sides. Anything subtracted from zero gives its negation.

$$-12x^{2}+25x=-7$$

Divide both sides by -12.

$$\frac{-12x^{2}+25x}{-12}=-\frac{7}{-12}$$

Dividing by -12 undoes the multiplication by -12.

$$x^{2}+\frac{25}{-12}x=-\frac{7}{-12}$$

Divide 25 by -12.

$$x^{2}-\frac{25}{12}x=-\frac{7}{-12}$$

Divide -7 by -12.

$$x^{2}-\frac{25}{12}x=\frac{7}{12}$$

Divide -\frac{25}{12}, the coefficient of the x term, by 2 to get -\frac{25}{24}. Then add the square of -\frac{25}{24} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{25}{12}x+\left(-\frac{25}{24}\right)^{2}=\frac{7}{12}+\left(-\frac{25}{24}\right)^{2}$$

Square -\frac{25}{24} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{25}{12}x+\frac{625}{576}=\frac{7}{12}+\frac{625}{576}$$

Add \frac{7}{12} to \frac{625}{576} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{25}{12}x+\frac{625}{576}=\frac{961}{576}$$

Factor x^{2}-\frac{25}{12}x+\frac{625}{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}.

$$\left(x-\frac{25}{24}\right)^{2}=\frac{961}{576}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{25}{24}\right)^{2}}=\sqrt{\frac{961}{576}}$$

Simplify.

$$x-\frac{25}{24}=\frac{31}{24} x-\frac{25}{24}=-\frac{31}{24}$$

Add \frac{25}{24} to both sides of the equation.

$$x=\frac{7}{3} x=-\frac{1}{4}$$