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

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

Solve for x

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

$$x=2$$

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.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

Use the distributive property to multiply x by x-1.

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

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

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

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}-9+x=5$$

Subtract 5 from both sides.

$$3x^{2}-9+x-5=0$$

Subtract 5 from -9 to get -14.

$$3x^{2}-14+x=0$$

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.

$$3x^{2}+x-14=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 1 for b, and -14 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\times 3\left(-14\right)}}{2\times 3}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\times 3\left(-14\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-1±\sqrt{1-12\left(-14\right)}}{2\times 3}$$

Multiply -12 times -14.

$$x=\frac{-1±\sqrt{1+168}}{2\times 3}$$

Add 1 to 168.

$$x=\frac{-1±\sqrt{169}}{2\times 3}$$

Take the square root of 169.

$$x=\frac{-1±13}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-1±13}{6}$$

Now solve the equation x=\frac{-1±13}{6} when ± is plus. Add -1 to 13.

$$x=\frac{12}{6}$$

Divide 12 by 6.

$$x=2$$

Now solve the equation x=\frac{-1±13}{6} when ± is minus. Subtract 13 from -1.

$$x=-\frac{14}{6}$$

Reduce the fraction \frac{-14}{6} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

$$x=2 x=-\frac{7}{3}$$

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.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

Use the distributive property to multiply x by x-1.

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

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

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

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}-9+x=5$$

Add 9 to both sides.

$$3x^{2}+x=5+9$$

Add 5 and 9 to get 14.

$$3x^{2}+x=14$$

Divide both sides by 3.

$$\frac{3x^{2}+x}{3}=\frac{14}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{1}{3}x=\frac{14}{3}$$

Divide \frac{1}{3}, the coefficient of the x term, by 2 to get \frac{1}{6}. Then add the square of \frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{3}x+\left(\frac{1}{6}\right)^{2}=\frac{14}{3}+\left(\frac{1}{6}\right)^{2}$$

Square \frac{1}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{3}x+\frac{1}{36}=\frac{14}{3}+\frac{1}{36}$$

Add \frac{14}{3} to \frac{1}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{3}x+\frac{1}{36}=\frac{169}{36}$$

Factor x^{2}+\frac{1}{3}x+\frac{1}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{6}\right)^{2}=\frac{169}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{6}\right)^{2}}=\sqrt{\frac{169}{36}}$$

Simplify.

$$x+\frac{1}{6}=\frac{13}{6} x+\frac{1}{6}=-\frac{13}{6}$$

Subtract \frac{1}{6} from both sides of the equation.

$$x=2 x=-\frac{7}{3}$$