Solve 3(2x+1)^2+10(2x+1)+3=0

$$3 ( 2 x + 1 ) ^ { 2 } + 10 ( 2 x + 1 ) + 3 = 0$$

Solve for x

$$x=-2$$

$$x=-\frac{2}{3}\approx -0.666666667$$

Steps Using Factoring By Grouping

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

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

Use the distributive property to multiply 3 by 4x^{2}+4x+1.

$$12x^{2}+12x+3+10\left(2x+1\right)+3=0$$

Use the distributive property to multiply 10 by 2x+1.

$$12x^{2}+12x+3+20x+10+3=0$$

Combine 12x and 20x to get 32x.

$$12x^{2}+32x+3+10+3=0$$

Add 3 and 10 to get 13.

$$12x^{2}+32x+13+3=0$$

Add 13 and 3 to get 16.

$$12x^{2}+32x+16=0$$

Divide both sides by 4.

$$3x^{2}+8x+4=0$$

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

$$a+b=8 ab=3\times 4=12$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 12.

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

Calculate the sum for each pair.

$$1+12=13 2+6=8 3+4=7$$

The solution is the pair that gives sum 8.

$$a=2 b=6$$

Rewrite 3x^{2}+8x+4 as \left(3x^{2}+2x\right)+\left(6x+4\right).

$$\left(3x^{2}+2x\right)+\left(6x+4\right)$$

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

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

Factor out common term 3x+2 by using distributive property.

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

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

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

Steps Using the Quadratic Formula

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

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

Use the distributive property to multiply 3 by 4x^{2}+4x+1.

$$12x^{2}+12x+3+10\left(2x+1\right)+3=0$$

Use the distributive property to multiply 10 by 2x+1.

$$12x^{2}+12x+3+20x+10+3=0$$

Combine 12x and 20x to get 32x.

$$12x^{2}+32x+3+10+3=0$$

Add 3 and 10 to get 13.

$$12x^{2}+32x+13+3=0$$

Add 13 and 3 to get 16.

$$12x^{2}+32x+16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 12 for a, 32 for b, and 16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-32±\sqrt{32^{2}-4\times 12\times 16}}{2\times 12}$$

Square 32.

$$x=\frac{-32±\sqrt{1024-4\times 12\times 16}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-32±\sqrt{1024-48\times 16}}{2\times 12}$$

Multiply -48 times 16.

$$x=\frac{-32±\sqrt{1024-768}}{2\times 12}$$

Add 1024 to -768.

$$x=\frac{-32±\sqrt{256}}{2\times 12}$$

Take the square root of 256.

$$x=\frac{-32±16}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{-32±16}{24}$$

Now solve the equation x=\frac{-32±16}{24} when ± is plus. Add -32 to 16.

$$x=-\frac{16}{24}$$

Reduce the fraction \frac{-16}{24} to lowest terms by extracting and canceling out 8.

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

Now solve the equation x=\frac{-32±16}{24} when ± is minus. Subtract 16 from -32.

$$x=-\frac{48}{24}$$

Divide -48 by 24.

$$x=-2$$

The equation is now solved.

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

Steps for Completing the Square

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

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

Use the distributive property to multiply 3 by 4x^{2}+4x+1.

$$12x^{2}+12x+3+10\left(2x+1\right)+3=0$$

Use the distributive property to multiply 10 by 2x+1.

$$12x^{2}+12x+3+20x+10+3=0$$

Combine 12x and 20x to get 32x.

$$12x^{2}+32x+3+10+3=0$$

Add 3 and 10 to get 13.

$$12x^{2}+32x+13+3=0$$

Add 13 and 3 to get 16.

$$12x^{2}+32x+16=0$$

Subtract 16 from both sides. Anything subtracted from zero gives its negation.

$$12x^{2}+32x=-16$$

Divide both sides by 12.

$$\frac{12x^{2}+32x}{12}=-\frac{16}{12}$$

Dividing by 12 undoes the multiplication by 12.

$$x^{2}+\frac{32}{12}x=-\frac{16}{12}$$

Reduce the fraction \frac{32}{12} to lowest terms by extracting and canceling out 4.

$$x^{2}+\frac{8}{3}x=-\frac{16}{12}$$

Reduce the fraction \frac{-16}{12} to lowest terms by extracting and canceling out 4.

$$x^{2}+\frac{8}{3}x=-\frac{4}{3}$$

Divide \frac{8}{3}, the coefficient of the x term, by 2 to get \frac{4}{3}. Then add the square of \frac{4}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{4}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{8}{3}x+\frac{16}{9}=-\frac{4}{3}+\frac{16}{9}$$

Add -\frac{4}{3} to \frac{16}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{8}{3}x+\frac{16}{9}=\frac{4}{9}$$

Factor x^{2}+\frac{8}{3}x+\frac{16}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{4}{3}\right)^{2}=\frac{4}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{4}{3}\right)^{2}}=\sqrt{\frac{4}{9}}$$

Simplify.

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

Subtract \frac{4}{3} from both sides of the equation.

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