Solve 3(x+4)^2+(x+4)-2=0

$$3 ( x + 4 ) ^ { 2 } + ( x + 4 ) - 2 = 0$$

Solve for x

$$x=-5$$

$$x = -\frac{10}{3} = -3\frac{1}{3} \approx -3.333333333$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+4\right)^{2}.

$$3\left(x^{2}+8x+16\right)+x+4-2=0$$

Use the distributive property to multiply 3 by x^{2}+8x+16.

$$3x^{2}+24x+48+x+4-2=0$$

Combine 24x and x to get 25x.

$$3x^{2}+25x+48+4-2=0$$

Add 48 and 4 to get 52.

$$3x^{2}+25x+52-2=0$$

Subtract 2 from 52 to get 50.

$$3x^{2}+25x+50=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx+50. To find a and b, set up a system to be solved.

$$a+b=25 ab=3\times 50=150$$

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 150.

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

Calculate the sum for each pair.

$$1+150=151 2+75=77 3+50=53 5+30=35 6+25=31 10+15=25$$

The solution is the pair that gives sum 25.

$$a=10 b=15$$

Rewrite 3x^{2}+25x+50 as \left(3x^{2}+10x\right)+\left(15x+50\right).

$$\left(3x^{2}+10x\right)+\left(15x+50\right)$$

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

$$x\left(3x+10\right)+5\left(3x+10\right)$$

Factor out common term 3x+10 by using distributive property.

$$\left(3x+10\right)\left(x+5\right)$$

To find equation solutions, solve 3x+10=0 and x+5=0.

$$x=-\frac{10}{3} x=-5$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+4\right)^{2}.

$$3\left(x^{2}+8x+16\right)+x+4-2=0$$

Use the distributive property to multiply 3 by x^{2}+8x+16.

$$3x^{2}+24x+48+x+4-2=0$$

Combine 24x and x to get 25x.

$$3x^{2}+25x+48+4-2=0$$

Add 48 and 4 to get 52.

$$3x^{2}+25x+52-2=0$$

Subtract 2 from 52 to get 50.

$$3x^{2}+25x+50=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 25 for b, and 50 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-25±\sqrt{25^{2}-4\times 3\times 50}}{2\times 3}$$

Square 25.

$$x=\frac{-25±\sqrt{625-4\times 3\times 50}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-25±\sqrt{625-12\times 50}}{2\times 3}$$

Multiply -12 times 50.

$$x=\frac{-25±\sqrt{625-600}}{2\times 3}$$

Add 625 to -600.

$$x=\frac{-25±\sqrt{25}}{2\times 3}$$

Take the square root of 25.

$$x=\frac{-25±5}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-25±5}{6}$$

Now solve the equation x=\frac{-25±5}{6} when ± is plus. Add -25 to 5.

$$x=-\frac{20}{6}$$

Reduce the fraction \frac{-20}{6} to lowest terms by extracting and canceling out 2.

$$x=-\frac{10}{3}$$

Now solve the equation x=\frac{-25±5}{6} when ± is minus. Subtract 5 from -25.

$$x=-\frac{30}{6}$$

Divide -30 by 6.

$$x=-5$$

The equation is now solved.

$$x=-\frac{10}{3} x=-5$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+4\right)^{2}.

$$3\left(x^{2}+8x+16\right)+x+4-2=0$$

Use the distributive property to multiply 3 by x^{2}+8x+16.

$$3x^{2}+24x+48+x+4-2=0$$

Combine 24x and x to get 25x.

$$3x^{2}+25x+48+4-2=0$$

Add 48 and 4 to get 52.

$$3x^{2}+25x+52-2=0$$

Subtract 2 from 52 to get 50.

$$3x^{2}+25x+50=0$$

Subtract 50 from both sides. Anything subtracted from zero gives its negation.

$$3x^{2}+25x=-50$$

Divide both sides by 3.

$$\frac{3x^{2}+25x}{3}=-\frac{50}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{25}{3}x=-\frac{50}{3}$$

Divide \frac{25}{3}, the coefficient of the x term, by 2 to get \frac{25}{6}. Then add the square of \frac{25}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{25}{3}x+\left(\frac{25}{6}\right)^{2}=-\frac{50}{3}+\left(\frac{25}{6}\right)^{2}$$

Square \frac{25}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{25}{3}x+\frac{625}{36}=-\frac{50}{3}+\frac{625}{36}$$

Add -\frac{50}{3} to \frac{625}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{25}{3}x+\frac{625}{36}=\frac{25}{36}$$

Factor x^{2}+\frac{25}{3}x+\frac{625}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{25}{6}\right)^{2}=\frac{25}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{25}{6}\right)^{2}}=\sqrt{\frac{25}{36}}$$

Simplify.

$$x+\frac{25}{6}=\frac{5}{6} x+\frac{25}{6}=-\frac{5}{6}$$

Subtract \frac{25}{6} from both sides of the equation.

$$x=-\frac{10}{3} x=-5$$