Solve -7x(x-1)=(x-1)(x+1)

$$-7x(x-1)=(x-1)(x+1)$$

Solve for x

$$x=1$$

$$x=-\frac{1}{8}=-0.125$$

Steps Using the Quadratic Formula

Use the distributive property to multiply -7x by x-1.

$$-7x^{2}+7x=\left(x-1\right)\left(x+1\right)$$

Consider \left(x-1\right)\left(x+1\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 1.

$$-7x^{2}+7x=x^{2}-1$$

Subtract x^{2} from both sides.

$$-7x^{2}+7x-x^{2}=-1$$

Combine -7x^{2} and -x^{2} to get -8x^{2}.

$$-8x^{2}+7x=-1$$

Add 1 to both sides.

$$-8x^{2}+7x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -8 for a, 7 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\left(-8\right)}}{2\left(-8\right)}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\left(-8\right)}}{2\left(-8\right)}$$

Multiply -4 times -8.

$$x=\frac{-7±\sqrt{49+32}}{2\left(-8\right)}$$

Add 49 to 32.

$$x=\frac{-7±\sqrt{81}}{2\left(-8\right)}$$

Take the square root of 81.

$$x=\frac{-7±9}{2\left(-8\right)}$$

Multiply 2 times -8.

$$x=\frac{-7±9}{-16}$$

Now solve the equation x=\frac{-7±9}{-16} when ± is plus. Add -7 to 9.

$$x=\frac{2}{-16}$$

Reduce the fraction \frac{2}{-16} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{-7±9}{-16} when ± is minus. Subtract 9 from -7.

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

Divide -16 by -16.

$$x=1$$

The equation is now solved.

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

Steps for Completing the Square

Use the distributive property to multiply -7x by x-1.

$$-7x^{2}+7x=\left(x-1\right)\left(x+1\right)$$

Consider \left(x-1\right)\left(x+1\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 1.

$$-7x^{2}+7x=x^{2}-1$$

Subtract x^{2} from both sides.

$$-7x^{2}+7x-x^{2}=-1$$

Combine -7x^{2} and -x^{2} to get -8x^{2}.

$$-8x^{2}+7x=-1$$

Divide both sides by -8.

$$\frac{-8x^{2}+7x}{-8}=-\frac{1}{-8}$$

Dividing by -8 undoes the multiplication by -8.

$$x^{2}+\frac{7}{-8}x=-\frac{1}{-8}$$

Divide 7 by -8.

$$x^{2}-\frac{7}{8}x=-\frac{1}{-8}$$

Divide -1 by -8.

$$x^{2}-\frac{7}{8}x=\frac{1}{8}$$

Divide -\frac{7}{8}, the coefficient of the x term, by 2 to get -\frac{7}{16}. Then add the square of -\frac{7}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{8}x+\left(-\frac{7}{16}\right)^{2}=\frac{1}{8}+\left(-\frac{7}{16}\right)^{2}$$

Square -\frac{7}{16} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{8}x+\frac{49}{256}=\frac{1}{8}+\frac{49}{256}$$

Add \frac{1}{8} to \frac{49}{256} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{7}{8}x+\frac{49}{256}=\frac{81}{256}$$

Factor x^{2}-\frac{7}{8}x+\frac{49}{256}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{7}{16}\right)^{2}=\frac{81}{256}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{16}\right)^{2}}=\sqrt{\frac{81}{256}}$$

Simplify.

$$x-\frac{7}{16}=\frac{9}{16} x-\frac{7}{16}=-\frac{9}{16}$$

Add \frac{7}{16} to both sides of the equation.

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