Solve (4x-1)(2x+3)=(3x+3)(x-1)

$$( 4 x - 1 ) ( 2 x + 3 ) = ( 3 x + 3 ) ( x - 1 )$$

Solve for x

$$x=-2$$

$$x=0$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 4x-1 by 2x+3 and combine like terms.

$$8x^{2}+10x-3=\left(3x+3\right)\left(x-1\right)$$

Use the distributive property to multiply 3x+3 by x-1 and combine like terms.

$$8x^{2}+10x-3=3x^{2}-3$$

Subtract 3x^{2} from both sides.

$$8x^{2}+10x-3-3x^{2}=-3$$

Combine 8x^{2} and -3x^{2} to get 5x^{2}.

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

Add 3 to both sides.

$$5x^{2}+10x-3+3=0$$

Add -3 and 3 to get 0.

$$5x^{2}+10x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 10 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}}}{2\times 5}$$

Take the square root of 10^{2}.

$$x=\frac{-10±10}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-10±10}{10}$$

Now solve the equation x=\frac{-10±10}{10} when ± is plus. Add -10 to 10.

$$x=\frac{0}{10}$$

Divide 0 by 10.

$$x=0$$

Now solve the equation x=\frac{-10±10}{10} when ± is minus. Subtract 10 from -10.

$$x=-\frac{20}{10}$$

Divide -20 by 10.

$$x=-2$$

The equation is now solved.

$$x=0 x=-2$$

Steps for Completing the Square

Use the distributive property to multiply 4x-1 by 2x+3 and combine like terms.

$$8x^{2}+10x-3=\left(3x+3\right)\left(x-1\right)$$

Use the distributive property to multiply 3x+3 by x-1 and combine like terms.

$$8x^{2}+10x-3=3x^{2}-3$$

Subtract 3x^{2} from both sides.

$$8x^{2}+10x-3-3x^{2}=-3$$

Combine 8x^{2} and -3x^{2} to get 5x^{2}.

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

Add 3 to both sides.

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

Add -3 and 3 to get 0.

$$5x^{2}+10x=0$$

Divide both sides by 5.

$$\frac{5x^{2}+10x}{5}=\frac{0}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}+\frac{10}{5}x=\frac{0}{5}$$

Divide 10 by 5.

$$x^{2}+2x=\frac{0}{5}$$

Divide 0 by 5.

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

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

$$x^{2}+2x+1^{2}=1^{2}$$

Square 1.

$$x^{2}+2x+1=1$$

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

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{1}$$

Simplify.

$$x+1=1 x+1=-1$$

Subtract 1 from both sides of the equation.

$$x=0 x=-2$$