Solve (4x-3)^2-2(x+3)=0

$$( 4 x - 3 ) ^ { 2 } - 2 ( x + 3 ) = 0$$

Solve for x

$$x=\frac{1}{8}=0.125$$

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4x-3\right)^{2}.

$$16x^{2}-24x+9-2\left(x+3\right)=0$$

Use the distributive property to multiply -2 by x+3.

$$16x^{2}-24x+9-2x-6=0$$

Combine -24x and -2x to get -26x.

$$16x^{2}-26x+9-6=0$$

Subtract 6 from 9 to get 3.

$$16x^{2}-26x+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 16x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=-26 ab=16\times 3=48$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 48.

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

Calculate the sum for each pair.

$$-1-48=-49 -2-24=-26 -3-16=-19 -4-12=-16 -6-8=-14$$

The solution is the pair that gives sum -26.

$$a=-24 b=-2$$

Rewrite 16x^{2}-26x+3 as \left(16x^{2}-24x\right)+\left(-2x+3\right).

$$\left(16x^{2}-24x\right)+\left(-2x+3\right)$$

Factor out 8x in the first and -1 in the second group.

$$8x\left(2x-3\right)-\left(2x-3\right)$$

Factor out common term 2x-3 by using distributive property.

$$\left(2x-3\right)\left(8x-1\right)$$

To find equation solutions, solve 2x-3=0 and 8x-1=0.

$$x=\frac{3}{2} x=\frac{1}{8}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4x-3\right)^{2}.

$$16x^{2}-24x+9-2\left(x+3\right)=0$$

Use the distributive property to multiply -2 by x+3.

$$16x^{2}-24x+9-2x-6=0$$

Combine -24x and -2x to get -26x.

$$16x^{2}-26x+9-6=0$$

Subtract 6 from 9 to get 3.

$$16x^{2}-26x+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, -26 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-26\right)±\sqrt{\left(-26\right)^{2}-4\times 16\times 3}}{2\times 16}$$

Square -26.

$$x=\frac{-\left(-26\right)±\sqrt{676-4\times 16\times 3}}{2\times 16}$$

Multiply -4 times 16.

$$x=\frac{-\left(-26\right)±\sqrt{676-64\times 3}}{2\times 16}$$

Multiply -64 times 3.

$$x=\frac{-\left(-26\right)±\sqrt{676-192}}{2\times 16}$$

Add 676 to -192.

$$x=\frac{-\left(-26\right)±\sqrt{484}}{2\times 16}$$

Take the square root of 484.

$$x=\frac{-\left(-26\right)±22}{2\times 16}$$

The opposite of -26 is 26.

$$x=\frac{26±22}{2\times 16}$$

Multiply 2 times 16.

$$x=\frac{26±22}{32}$$

Now solve the equation x=\frac{26±22}{32} when ± is plus. Add 26 to 22.

$$x=\frac{48}{32}$$

Reduce the fraction \frac{48}{32} to lowest terms by extracting and canceling out 16.

$$x=\frac{3}{2}$$

Now solve the equation x=\frac{26±22}{32} when ± is minus. Subtract 22 from 26.

$$x=\frac{4}{32}$$

Reduce the fraction \frac{4}{32} to lowest terms by extracting and canceling out 4.

$$x=\frac{1}{8}$$

The equation is now solved.

$$x=\frac{3}{2} x=\frac{1}{8}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4x-3\right)^{2}.

$$16x^{2}-24x+9-2\left(x+3\right)=0$$

Use the distributive property to multiply -2 by x+3.

$$16x^{2}-24x+9-2x-6=0$$

Combine -24x and -2x to get -26x.

$$16x^{2}-26x+9-6=0$$

Subtract 6 from 9 to get 3.

$$16x^{2}-26x+3=0$$

Subtract 3 from both sides. Anything subtracted from zero gives its negation.

$$16x^{2}-26x=-3$$

Divide both sides by 16.

$$\frac{16x^{2}-26x}{16}=-\frac{3}{16}$$

Dividing by 16 undoes the multiplication by 16.

$$x^{2}+\left(-\frac{26}{16}\right)x=-\frac{3}{16}$$

Reduce the fraction \frac{-26}{16} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{13}{8}x=-\frac{3}{16}$$

Divide -\frac{13}{8}, the coefficient of the x term, by 2 to get -\frac{13}{16}. Then add the square of -\frac{13}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{13}{8}x+\left(-\frac{13}{16}\right)^{2}=-\frac{3}{16}+\left(-\frac{13}{16}\right)^{2}$$

Square -\frac{13}{16} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{13}{8}x+\frac{169}{256}=-\frac{3}{16}+\frac{169}{256}$$

Add -\frac{3}{16} to \frac{169}{256} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{13}{8}x+\frac{169}{256}=\frac{121}{256}$$

Factor x^{2}-\frac{13}{8}x+\frac{169}{256}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{13}{16}\right)^{2}=\frac{121}{256}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{13}{16}\right)^{2}}=\sqrt{\frac{121}{256}}$$

Simplify.

$$x-\frac{13}{16}=\frac{11}{16} x-\frac{13}{16}=-\frac{11}{16}$$

Add \frac{13}{16} to both sides of the equation.

$$x=\frac{3}{2} x=\frac{1}{8}$$