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

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

Solve for x

$$x=-\frac{1}{5}=-0.2$$

$$x=0$$

Steps Using the Quadratic Formula

Consider \left(5x-1\right)\left(5x+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.

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

Expand \left(5x\right)^{2}.

$$5^{2}x^{2}-1=-1-5x$$

Calculate 5 to the power of 2 and get 25.

$$25x^{2}-1=-1-5x$$

Subtract -1 from both sides.

$$25x^{2}-1-\left(-1\right)=-5x$$

The opposite of -1 is 1.

$$25x^{2}-1+1=-5x$$

Add 5x to both sides.

$$25x^{2}-1+1+5x=0$$

Add -1 and 1 to get 0.

$$25x^{2}+5x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 25 for a, 5 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}}}{2\times 25}$$

Take the square root of 5^{2}.

$$x=\frac{-5±5}{2\times 25}$$

Multiply 2 times 25.

$$x=\frac{-5±5}{50}$$

Now solve the equation x=\frac{-5±5}{50} when ± is plus. Add -5 to 5.

$$x=\frac{0}{50}$$

Divide 0 by 50.

$$x=0$$

Now solve the equation x=\frac{-5±5}{50} when ± is minus. Subtract 5 from -5.

$$x=-\frac{10}{50}$$

Reduce the fraction \frac{-10}{50} to lowest terms by extracting and canceling out 10.

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

The equation is now solved.

$$x=0 x=-\frac{1}{5}$$

Steps for Completing the Square

Consider \left(5x-1\right)\left(5x+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.

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

Expand \left(5x\right)^{2}.

$$5^{2}x^{2}-1=-1-5x$$

Calculate 5 to the power of 2 and get 25.

$$25x^{2}-1=-1-5x$$

Add 5x to both sides.

$$25x^{2}-1+5x=-1$$

Add 1 to both sides.

$$25x^{2}+5x=-1+1$$

Add -1 and 1 to get 0.

$$25x^{2}+5x=0$$

Divide both sides by 25.

$$\frac{25x^{2}+5x}{25}=\frac{0}{25}$$

Dividing by 25 undoes the multiplication by 25.

$$x^{2}+\frac{5}{25}x=\frac{0}{25}$$

Reduce the fraction \frac{5}{25} to lowest terms by extracting and canceling out 5.

$$x^{2}+\frac{1}{5}x=\frac{0}{25}$$

Divide 0 by 25.

$$x^{2}+\frac{1}{5}x=0$$

Divide \frac{1}{5}, the coefficient of the x term, by 2 to get \frac{1}{10}. Then add the square of \frac{1}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{1}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{5}x+\frac{1}{100}=\frac{1}{100}$$

Factor x^{2}+\frac{1}{5}x+\frac{1}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{10}\right)^{2}=\frac{1}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{10}\right)^{2}}=\sqrt{\frac{1}{100}}$$

Simplify.

$$x+\frac{1}{10}=\frac{1}{10} x+\frac{1}{10}=-\frac{1}{10}$$

Subtract \frac{1}{10} from both sides of the equation.

$$x=0 x=-\frac{1}{5}$$