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

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

Solve for x

$$x=-\frac{3}{4}=-0.75$$

$$x=1$$

Steps Using Factoring By Grouping

Use the distributive property to multiply x-1 by 5x+3 and combine like terms.

$$5x^{2}-2x-3=x\left(x-1\right)$$

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

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

Subtract x^{2} from both sides.

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

Combine 5x^{2} and -x^{2} to get 4x^{2}.

$$4x^{2}-2x-3=-x$$

Add x to both sides.

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

Combine -2x and x to get -x.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=-1 ab=4\left(-3\right)=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -12.

$$1,-12 2,-6 3,-4$$

Calculate the sum for each pair.

$$1-12=-11 2-6=-4 3-4=-1$$

The solution is the pair that gives sum -1.

$$a=-4 b=3$$

Rewrite 4x^{2}-x-3 as \left(4x^{2}-4x\right)+\left(3x-3\right).

$$\left(4x^{2}-4x\right)+\left(3x-3\right)$$

Factor out 4x in the first and 3 in the second group.

$$4x\left(x-1\right)+3\left(x-1\right)$$

Factor out common term x-1 by using distributive property.

$$\left(x-1\right)\left(4x+3\right)$$

To find equation solutions, solve x-1=0 and 4x+3=0.

$$x=1 x=-\frac{3}{4}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x-1 by 5x+3 and combine like terms.

$$5x^{2}-2x-3=x\left(x-1\right)$$

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

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

Subtract x^{2} from both sides.

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

Combine 5x^{2} and -x^{2} to get 4x^{2}.

$$4x^{2}-2x-3=-x$$

Add x to both sides.

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

Combine -2x and x to get -x.

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

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

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

Multiply -4 times 4.

$$x=\frac{-\left(-1\right)±\sqrt{1-16\left(-3\right)}}{2\times 4}$$

Multiply -16 times -3.

$$x=\frac{-\left(-1\right)±\sqrt{1+48}}{2\times 4}$$

Add 1 to 48.

$$x=\frac{-\left(-1\right)±\sqrt{49}}{2\times 4}$$

Take the square root of 49.

$$x=\frac{-\left(-1\right)±7}{2\times 4}$$

The opposite of -1 is 1.

$$x=\frac{1±7}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{1±7}{8}$$

Now solve the equation x=\frac{1±7}{8} when ± is plus. Add 1 to 7.

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

Divide 8 by 8.

$$x=1$$

Now solve the equation x=\frac{1±7}{8} when ± is minus. Subtract 7 from 1.

$$x=-\frac{6}{8}$$

Reduce the fraction \frac{-6}{8} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

$$x=1 x=-\frac{3}{4}$$

Steps for Completing the Square

Use the distributive property to multiply x-1 by 5x+3 and combine like terms.

$$5x^{2}-2x-3=x\left(x-1\right)$$

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

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

Subtract x^{2} from both sides.

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

Combine 5x^{2} and -x^{2} to get 4x^{2}.

$$4x^{2}-2x-3=-x$$

Add x to both sides.

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

Combine -2x and x to get -x.

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

Add 3 to both sides. Anything plus zero gives itself.

$$4x^{2}-x=3$$

Divide both sides by 4.

$$\frac{4x^{2}-x}{4}=\frac{3}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}-\frac{1}{4}x=\frac{3}{4}$$

Divide -\frac{1}{4}, the coefficient of the x term, by 2 to get -\frac{1}{8}. Then add the square of -\frac{1}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{4}x+\left(-\frac{1}{8}\right)^{2}=\frac{3}{4}+\left(-\frac{1}{8}\right)^{2}$$

Square -\frac{1}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{4}x+\frac{1}{64}=\frac{3}{4}+\frac{1}{64}$$

Add \frac{3}{4} to \frac{1}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{4}x+\frac{1}{64}=\frac{49}{64}$$

Factor x^{2}-\frac{1}{4}x+\frac{1}{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}.

$$\left(x-\frac{1}{8}\right)^{2}=\frac{49}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{8}\right)^{2}}=\sqrt{\frac{49}{64}}$$

Simplify.

$$x-\frac{1}{8}=\frac{7}{8} x-\frac{1}{8}=-\frac{7}{8}$$

Add \frac{1}{8} to both sides of the equation.

$$x=1 x=-\frac{3}{4}$$