Solve (5x+3)^2=5(x+3)

$$( 5 x + 3 ) ^ { 2 } = 5 ( x + 3 )$$

Solve for x

$$x = -\frac{6}{5} = -1\frac{1}{5} = -1.2$$

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

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(5x+3\right)^{2}.

$$25x^{2}+30x+9=5\left(x+3\right)$$

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

$$25x^{2}+30x+9=5x+15$$

Subtract 5x from both sides.

$$25x^{2}+30x+9-5x=15$$

Combine 30x and -5x to get 25x.

$$25x^{2}+25x+9=15$$

Subtract 15 from both sides.

$$25x^{2}+25x+9-15=0$$

Subtract 15 from 9 to get -6.

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

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.

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

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. 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 25.

$$a=-5 b=30$$

Rewrite 25x^{2}+25x-6 as \left(25x^{2}-5x\right)+\left(30x-6\right).

$$\left(25x^{2}-5x\right)+\left(30x-6\right)$$

Factor out 5x in the first and 6 in the second group.

$$5x\left(5x-1\right)+6\left(5x-1\right)$$

Factor out common term 5x-1 by using distributive property.

$$\left(5x-1\right)\left(5x+6\right)$$

To find equation solutions, solve 5x-1=0 and 5x+6=0.

$$x=\frac{1}{5} x=-\frac{6}{5}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(5x+3\right)^{2}.

$$25x^{2}+30x+9=5\left(x+3\right)$$

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

$$25x^{2}+30x+9=5x+15$$

Subtract 5x from both sides.

$$25x^{2}+30x+9-5x=15$$

Combine 30x and -5x to get 25x.

$$25x^{2}+25x+9=15$$

Subtract 15 from both sides.

$$25x^{2}+25x+9-15=0$$

Subtract 15 from 9 to get -6.

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

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

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

Square 25.

$$x=\frac{-25±\sqrt{625-4\times 25\left(-6\right)}}{2\times 25}$$

Multiply -4 times 25.

$$x=\frac{-25±\sqrt{625-100\left(-6\right)}}{2\times 25}$$

Multiply -100 times -6.

$$x=\frac{-25±\sqrt{625+600}}{2\times 25}$$

Add 625 to 600.

$$x=\frac{-25±\sqrt{1225}}{2\times 25}$$

Take the square root of 1225.

$$x=\frac{-25±35}{2\times 25}$$

Multiply 2 times 25.

$$x=\frac{-25±35}{50}$$

Now solve the equation x=\frac{-25±35}{50} when ± is plus. Add -25 to 35.

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

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

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

Now solve the equation x=\frac{-25±35}{50} when ± is minus. Subtract 35 from -25.

$$x=-\frac{60}{50}$$

Reduce the fraction \frac{-60}{50} to lowest terms by extracting and canceling out 10.

$$x=-\frac{6}{5}$$

The equation is now solved.

$$x=\frac{1}{5} x=-\frac{6}{5}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(5x+3\right)^{2}.

$$25x^{2}+30x+9=5\left(x+3\right)$$

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

$$25x^{2}+30x+9=5x+15$$

Subtract 5x from both sides.

$$25x^{2}+30x+9-5x=15$$

Combine 30x and -5x to get 25x.

$$25x^{2}+25x+9=15$$

Subtract 9 from both sides.

$$25x^{2}+25x=15-9$$

Subtract 9 from 15 to get 6.

$$25x^{2}+25x=6$$

Divide both sides by 25.

$$\frac{25x^{2}+25x}{25}=\frac{6}{25}$$

Dividing by 25 undoes the multiplication by 25.

$$x^{2}+\frac{25}{25}x=\frac{6}{25}$$

Divide 25 by 25.

$$x^{2}+x=\frac{6}{25}$$

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

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

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+x+\frac{1}{4}=\frac{6}{25}+\frac{1}{4}$$

Add \frac{6}{25} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+x+\frac{1}{4}=\frac{49}{100}$$

Factor x^{2}+x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{49}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{49}{100}}$$

Simplify.

$$x+\frac{1}{2}=\frac{7}{10} x+\frac{1}{2}=-\frac{7}{10}$$

Subtract \frac{1}{2} from both sides of the equation.

$$x=\frac{1}{5} x=-\frac{6}{5}$$