Solve (2x+3)(2x-3)+5x=2(x+1)-1

$$( 2 x + 3 ) ( 2 x - 3 ) + 5 x = 2 ( x + 1 ) - 1$$

Solve for x

$$x=-2$$

$$x = \frac{5}{4} = 1\frac{1}{4} = 1.25$$

Steps Using the Quadratic Formula

Consider \left(2x+3\right)\left(2x-3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

$$\left(2x\right)^{2}-9+5x=2\left(x+1\right)-1$$

Expand \left(2x\right)^{2}.

$$2^{2}x^{2}-9+5x=2\left(x+1\right)-1$$

Calculate 2 to the power of 2 and get 4.

$$4x^{2}-9+5x=2\left(x+1\right)-1$$

Use the distributive property to multiply 2 by x+1.

$$4x^{2}-9+5x=2x+2-1$$

Subtract 1 from 2 to get 1.

$$4x^{2}-9+5x=2x+1$$

Subtract 2x from both sides.

$$4x^{2}-9+5x-2x=1$$

Combine 5x and -2x to get 3x.

$$4x^{2}-9+3x=1$$

Subtract 1 from both sides.

$$4x^{2}-9+3x-1=0$$

Subtract 1 from -9 to get -10.

$$4x^{2}-10+3x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$4x^{2}+3x-10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 3 for b, and -10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\times 4\left(-10\right)}}{2\times 4}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\times 4\left(-10\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-3±\sqrt{9-16\left(-10\right)}}{2\times 4}$$

Multiply -16 times -10.

$$x=\frac{-3±\sqrt{9+160}}{2\times 4}$$

Add 9 to 160.

$$x=\frac{-3±\sqrt{169}}{2\times 4}$$

Take the square root of 169.

$$x=\frac{-3±13}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-3±13}{8}$$

Now solve the equation x=\frac{-3±13}{8} when ± is plus. Add -3 to 13.

$$x=\frac{10}{8}$$

Reduce the fraction \frac{10}{8} to lowest terms by extracting and canceling out 2.

$$x=\frac{5}{4}$$

Now solve the equation x=\frac{-3±13}{8} when ± is minus. Subtract 13 from -3.

$$x=-\frac{16}{8}$$

Divide -16 by 8.

$$x=-2$$

The equation is now solved.

$$x=\frac{5}{4} x=-2$$

Steps for Completing the Square

Consider \left(2x+3\right)\left(2x-3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

$$\left(2x\right)^{2}-9+5x=2\left(x+1\right)-1$$

Expand \left(2x\right)^{2}.

$$2^{2}x^{2}-9+5x=2\left(x+1\right)-1$$

Calculate 2 to the power of 2 and get 4.

$$4x^{2}-9+5x=2\left(x+1\right)-1$$

Use the distributive property to multiply 2 by x+1.

$$4x^{2}-9+5x=2x+2-1$$

Subtract 1 from 2 to get 1.

$$4x^{2}-9+5x=2x+1$$

Subtract 2x from both sides.

$$4x^{2}-9+5x-2x=1$$

Combine 5x and -2x to get 3x.

$$4x^{2}-9+3x=1$$

Add 9 to both sides.

$$4x^{2}+3x=1+9$$

Add 1 and 9 to get 10.

$$4x^{2}+3x=10$$

Divide both sides by 4.

$$\frac{4x^{2}+3x}{4}=\frac{10}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{3}{4}x=\frac{10}{4}$$

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

$$x^{2}+\frac{3}{4}x=\frac{5}{2}$$

Divide \frac{3}{4}, the coefficient of the x term, by 2 to get \frac{3}{8}. Then add the square of \frac{3}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{3}{4}x+\left(\frac{3}{8}\right)^{2}=\frac{5}{2}+\left(\frac{3}{8}\right)^{2}$$

Square \frac{3}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{3}{4}x+\frac{9}{64}=\frac{5}{2}+\frac{9}{64}$$

Add \frac{5}{2} to \frac{9}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{3}{4}x+\frac{9}{64}=\frac{169}{64}$$

Factor x^{2}+\frac{3}{4}x+\frac{9}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{8}\right)^{2}=\frac{169}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{8}\right)^{2}}=\sqrt{\frac{169}{64}}$$

Simplify.

$$x+\frac{3}{8}=\frac{13}{8} x+\frac{3}{8}=-\frac{13}{8}$$

Subtract \frac{3}{8} from both sides of the equation.

$$x=\frac{5}{4} x=-2$$