Solve left(5x+1right)^2-3(5x+1)-4=0

$${ \left(5x+1 \right) }^{ 2 } -3(5x+1)-4=0$$

Solve for x

$$x=-\frac{2}{5}=-0.4$$

$$x=\frac{3}{5}=0.6$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(5x+1\right)^{2}.

$$25x^{2}+10x+1-3\left(5x+1\right)-4=0$$

Use the distributive property to multiply -3 by 5x+1.

$$25x^{2}+10x+1-15x-3-4=0$$

Combine 10x and -15x to get -5x.

$$25x^{2}-5x+1-3-4=0$$

Subtract 3 from 1 to get -2.

$$25x^{2}-5x-2-4=0$$

Subtract 4 from -2 to get -6.

$$25x^{2}-5x-6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 25x^{2}+ax+bx-6. To find a and b, set up a system to be solved.

$$a+b=-5 ab=25\left(-6\right)=-150$$

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 -150.

$$1,-150 2,-75 3,-50 5,-30 6,-25 10,-15$$

Calculate the sum for each pair.

$$1-150=-149 2-75=-73 3-50=-47 5-30=-25 6-25=-19 10-15=-5$$

The solution is the pair that gives sum -5.

$$a=-15 b=10$$

Rewrite 25x^{2}-5x-6 as \left(25x^{2}-15x\right)+\left(10x-6\right).

$$\left(25x^{2}-15x\right)+\left(10x-6\right)$$

Factor out 5x in the first and 2 in the second group.

$$5x\left(5x-3\right)+2\left(5x-3\right)$$

Factor out common term 5x-3 by using distributive property.

$$\left(5x-3\right)\left(5x+2\right)$$

To find equation solutions, solve 5x-3=0 and 5x+2=0.

$$x=\frac{3}{5} x=-\frac{2}{5}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(5x+1\right)^{2}.

$$25x^{2}+10x+1-3\left(5x+1\right)-4=0$$

Use the distributive property to multiply -3 by 5x+1.

$$25x^{2}+10x+1-15x-3-4=0$$

Combine 10x and -15x to get -5x.

$$25x^{2}-5x+1-3-4=0$$

Subtract 3 from 1 to get -2.

$$25x^{2}-5x-2-4=0$$

Subtract 4 from -2 to get -6.

$$25x^{2}-5x-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 25 for a, -5 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 25\left(-6\right)}}{2\times 25}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 25\left(-6\right)}}{2\times 25}$$

Multiply -4 times 25.

$$x=\frac{-\left(-5\right)±\sqrt{25-100\left(-6\right)}}{2\times 25}$$

Multiply -100 times -6.

$$x=\frac{-\left(-5\right)±\sqrt{25+600}}{2\times 25}$$

Add 25 to 600.

$$x=\frac{-\left(-5\right)±\sqrt{625}}{2\times 25}$$

Take the square root of 625.

$$x=\frac{-\left(-5\right)±25}{2\times 25}$$

The opposite of -5 is 5.

$$x=\frac{5±25}{2\times 25}$$

Multiply 2 times 25.

$$x=\frac{5±25}{50}$$

Now solve the equation x=\frac{5±25}{50} when ± is plus. Add 5 to 25.

$$x=\frac{30}{50}$$

Reduce the fraction \frac{30}{50} to lowest terms by extracting and canceling out 10.

$$x=\frac{3}{5}$$

Now solve the equation x=\frac{5±25}{50} when ± is minus. Subtract 25 from 5.

$$x=-\frac{20}{50}$$

Reduce the fraction \frac{-20}{50} to lowest terms by extracting and canceling out 10.

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

The equation is now solved.

$$x=\frac{3}{5} x=-\frac{2}{5}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(5x+1\right)^{2}.

$$25x^{2}+10x+1-3\left(5x+1\right)-4=0$$

Use the distributive property to multiply -3 by 5x+1.

$$25x^{2}+10x+1-15x-3-4=0$$

Combine 10x and -15x to get -5x.

$$25x^{2}-5x+1-3-4=0$$

Subtract 3 from 1 to get -2.

$$25x^{2}-5x-2-4=0$$

Subtract 4 from -2 to get -6.

$$25x^{2}-5x-6=0$$

Add 6 to both sides. Anything plus zero gives itself.

$$25x^{2}-5x=6$$

Divide both sides by 25.

$$\frac{25x^{2}-5x}{25}=\frac{6}{25}$$

Dividing by 25 undoes the multiplication by 25.

$$x^{2}+\left(-\frac{5}{25}\right)x=\frac{6}{25}$$

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

$$x^{2}-\frac{1}{5}x=\frac{6}{25}$$

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.

$$x^{2}-\frac{1}{5}x+\left(-\frac{1}{10}\right)^{2}=\frac{6}{25}+\left(-\frac{1}{10}\right)^{2}$$

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

$$x^{2}-\frac{1}{5}x+\frac{1}{100}=\frac{6}{25}+\frac{1}{100}$$

Add \frac{6}{25} to \frac{1}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{5}x+\frac{1}{100}=\frac{1}{4}$$

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-\frac{1}{10}=\frac{1}{2} x-\frac{1}{10}=-\frac{1}{2}$$

Add \frac{1}{10} to both sides of the equation.

$$x=\frac{3}{5} x=-\frac{2}{5}$$