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

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

Solve for x

$$x = -\frac{5}{2} = -2\frac{1}{2} = -2.5$$

$$x=-1$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2x by 2x+5.

$$4x^{2}+10x=\left(x-1\right)\left(2x+5\right)$$

Use the distributive property to multiply x-1 by 2x+5 and combine like terms.

$$4x^{2}+10x=2x^{2}+3x-5$$

Subtract 2x^{2} from both sides.

$$4x^{2}+10x-2x^{2}=3x-5$$

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

$$2x^{2}+10x=3x-5$$

Subtract 3x from both sides.

$$2x^{2}+10x-3x=-5$$

Combine 10x and -3x to get 7x.

$$2x^{2}+7x=-5$$

Add 5 to both sides.

$$2x^{2}+7x+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 7 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\times 2\times 5}}{2\times 2}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\times 2\times 5}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-7±\sqrt{49-8\times 5}}{2\times 2}$$

Multiply -8 times 5.

$$x=\frac{-7±\sqrt{49-40}}{2\times 2}$$

Add 49 to -40.

$$x=\frac{-7±\sqrt{9}}{2\times 2}$$

Take the square root of 9.

$$x=\frac{-7±3}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-7±3}{4}$$

Now solve the equation x=\frac{-7±3}{4} when ± is plus. Add -7 to 3.

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

Divide -4 by 4.

$$x=-1$$

Now solve the equation x=\frac{-7±3}{4} when ± is minus. Subtract 3 from -7.

$$x=-\frac{10}{4}$$

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

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

The equation is now solved.

$$x=-1 x=-\frac{5}{2}$$

Steps for Completing the Square

Use the distributive property to multiply 2x by 2x+5.

$$4x^{2}+10x=\left(x-1\right)\left(2x+5\right)$$

Use the distributive property to multiply x-1 by 2x+5 and combine like terms.

$$4x^{2}+10x=2x^{2}+3x-5$$

Subtract 2x^{2} from both sides.

$$4x^{2}+10x-2x^{2}=3x-5$$

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

$$2x^{2}+10x=3x-5$$

Subtract 3x from both sides.

$$2x^{2}+10x-3x=-5$$

Combine 10x and -3x to get 7x.

$$2x^{2}+7x=-5$$

Divide both sides by 2.

$$\frac{2x^{2}+7x}{2}=-\frac{5}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{7}{2}x=-\frac{5}{2}$$

Divide \frac{7}{2}, the coefficient of the x term, by 2 to get \frac{7}{4}. Then add the square of \frac{7}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{7}{2}x+\left(\frac{7}{4}\right)^{2}=-\frac{5}{2}+\left(\frac{7}{4}\right)^{2}$$

Square \frac{7}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{7}{2}x+\frac{49}{16}=-\frac{5}{2}+\frac{49}{16}$$

Add -\frac{5}{2} to \frac{49}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{7}{2}x+\frac{49}{16}=\frac{9}{16}$$

Factor x^{2}+\frac{7}{2}x+\frac{49}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{7}{4}\right)^{2}=\frac{9}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Simplify.

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

Subtract \frac{7}{4} from both sides of the equation.

$$x=-1 x=-\frac{5}{2}$$