Solve 5x+6=left(3x+2right)^2

$$5x+6 = { \left(3x+2 \right) }^{ 2 }$$

Solve for x

$$x=-1$$

$$x=\frac{2}{9}\approx 0.222222222$$

Steps Using Factoring By Grouping

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

$$5x+6=9x^{2}+12x+4$$

Subtract 9x^{2} from both sides.

$$5x+6-9x^{2}=12x+4$$

Subtract 12x from both sides.

$$5x+6-9x^{2}-12x=4$$

Combine 5x and -12x to get -7x.

$$-7x+6-9x^{2}=4$$

Subtract 4 from both sides.

$$-7x+6-9x^{2}-4=0$$

Subtract 4 from 6 to get 2.

$$-7x+2-9x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-9x^{2}-7x+2=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -9x^{2}+ax+bx+2. To find a and b, set up a system to be solved.

$$a+b=-7 ab=-9\times 2=-18$$

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 -18.

$$1,-18 2,-9 3,-6$$

Calculate the sum for each pair.

$$1-18=-17 2-9=-7 3-6=-3$$

The solution is the pair that gives sum -7.

$$a=2 b=-9$$

Rewrite -9x^{2}-7x+2 as \left(-9x^{2}+2x\right)+\left(-9x+2\right).

$$\left(-9x^{2}+2x\right)+\left(-9x+2\right)$$

Factor out -x in the first and -1 in the second group.

$$-x\left(9x-2\right)-\left(9x-2\right)$$

Factor out common term 9x-2 by using distributive property.

$$\left(9x-2\right)\left(-x-1\right)$$

To find equation solutions, solve 9x-2=0 and -x-1=0.

$$x=\frac{2}{9} x=-1$$

Steps Using the Quadratic Formula

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

$$5x+6=9x^{2}+12x+4$$

Subtract 9x^{2} from both sides.

$$5x+6-9x^{2}=12x+4$$

Subtract 12x from both sides.

$$5x+6-9x^{2}-12x=4$$

Combine 5x and -12x to get -7x.

$$-7x+6-9x^{2}=4$$

Subtract 4 from both sides.

$$-7x+6-9x^{2}-4=0$$

Subtract 4 from 6 to get 2.

$$-7x+2-9x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-9x^{2}-7x+2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -9 for a, -7 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\left(-9\right)\times 2}}{2\left(-9\right)}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\left(-9\right)\times 2}}{2\left(-9\right)}$$

Multiply -4 times -9.

$$x=\frac{-\left(-7\right)±\sqrt{49+36\times 2}}{2\left(-9\right)}$$

Multiply 36 times 2.

$$x=\frac{-\left(-7\right)±\sqrt{49+72}}{2\left(-9\right)}$$

Add 49 to 72.

$$x=\frac{-\left(-7\right)±\sqrt{121}}{2\left(-9\right)}$$

Take the square root of 121.

$$x=\frac{-\left(-7\right)±11}{2\left(-9\right)}$$

The opposite of -7 is 7.

$$x=\frac{7±11}{2\left(-9\right)}$$

Multiply 2 times -9.

$$x=\frac{7±11}{-18}$$

Now solve the equation x=\frac{7±11}{-18} when ± is plus. Add 7 to 11.

$$x=\frac{18}{-18}$$

Divide 18 by -18.

$$x=-1$$

Now solve the equation x=\frac{7±11}{-18} when ± is minus. Subtract 11 from 7.

$$x=-\frac{4}{-18}$$

Reduce the fraction \frac{-4}{-18} to lowest terms by extracting and canceling out 2.

$$x=\frac{2}{9}$$

The equation is now solved.

$$x=-1 x=\frac{2}{9}$$

Steps for Completing the Square

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

$$5x+6=9x^{2}+12x+4$$

Subtract 9x^{2} from both sides.

$$5x+6-9x^{2}=12x+4$$

Subtract 12x from both sides.

$$5x+6-9x^{2}-12x=4$$

Combine 5x and -12x to get -7x.

$$-7x+6-9x^{2}=4$$

Subtract 6 from both sides.

$$-7x-9x^{2}=4-6$$

Subtract 6 from 4 to get -2.

$$-7x-9x^{2}=-2$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-9x^{2}-7x=-2$$

Divide both sides by -9.

$$\frac{-9x^{2}-7x}{-9}=-\frac{2}{-9}$$

Dividing by -9 undoes the multiplication by -9.

$$x^{2}+\left(-\frac{7}{-9}\right)x=-\frac{2}{-9}$$

Divide -7 by -9.

$$x^{2}+\frac{7}{9}x=-\frac{2}{-9}$$

Divide -2 by -9.

$$x^{2}+\frac{7}{9}x=\frac{2}{9}$$

Divide \frac{7}{9}, the coefficient of the x term, by 2 to get \frac{7}{18}. Then add the square of \frac{7}{18} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{7}{9}x+\left(\frac{7}{18}\right)^{2}=\frac{2}{9}+\left(\frac{7}{18}\right)^{2}$$

Square \frac{7}{18} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{7}{9}x+\frac{49}{324}=\frac{2}{9}+\frac{49}{324}$$

Add \frac{2}{9} to \frac{49}{324} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{7}{9}x+\frac{49}{324}=\frac{121}{324}$$

Factor x^{2}+\frac{7}{9}x+\frac{49}{324}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{7}{18}\right)^{2}=\frac{121}{324}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{18}\right)^{2}}=\sqrt{\frac{121}{324}}$$

Simplify.

$$x+\frac{7}{18}=\frac{11}{18} x+\frac{7}{18}=-\frac{11}{18}$$

Subtract \frac{7}{18} from both sides of the equation.

$$x=\frac{2}{9} x=-1$$