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

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

Solve for x

$$x = \frac{9}{7} = 1\frac{2}{7} \approx 1.285714286$$

$$x=3$$

Steps Using Factoring By Grouping

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

$$3\left(x^{2}-6x+9\right)+4x\left(x-3\right)=0$$

Use the distributive property to multiply 3 by x^{2}-6x+9.

$$3x^{2}-18x+27+4x\left(x-3\right)=0$$

Use the distributive property to multiply 4x by x-3.

$$3x^{2}-18x+27+4x^{2}-12x=0$$

Combine 3x^{2} and 4x^{2} to get 7x^{2}.

$$7x^{2}-18x+27-12x=0$$

Combine -18x and -12x to get -30x.

$$7x^{2}-30x+27=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 7x^{2}+ax+bx+27. To find a and b, set up a system to be solved.

$$a+b=-30 ab=7\times 27=189$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 189.

$$-1,-189 -3,-63 -7,-27 -9,-21$$

Calculate the sum for each pair.

$$-1-189=-190 -3-63=-66 -7-27=-34 -9-21=-30$$

The solution is the pair that gives sum -30.

$$a=-21 b=-9$$

Rewrite 7x^{2}-30x+27 as \left(7x^{2}-21x\right)+\left(-9x+27\right).

$$\left(7x^{2}-21x\right)+\left(-9x+27\right)$$

Factor out 7x in the first and -9 in the second group.

$$7x\left(x-3\right)-9\left(x-3\right)$$

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

$$\left(x-3\right)\left(7x-9\right)$$

To find equation solutions, solve x-3=0 and 7x-9=0.

$$x=3 x=\frac{9}{7}$$

Steps Using the Quadratic Formula

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

$$3\left(x^{2}-6x+9\right)+4x\left(x-3\right)=0$$

Use the distributive property to multiply 3 by x^{2}-6x+9.

$$3x^{2}-18x+27+4x\left(x-3\right)=0$$

Use the distributive property to multiply 4x by x-3.

$$3x^{2}-18x+27+4x^{2}-12x=0$$

Combine 3x^{2} and 4x^{2} to get 7x^{2}.

$$7x^{2}-18x+27-12x=0$$

Combine -18x and -12x to get -30x.

$$7x^{2}-30x+27=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 7 for a, -30 for b, and 27 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-30\right)±\sqrt{\left(-30\right)^{2}-4\times 7\times 27}}{2\times 7}$$

Square -30.

$$x=\frac{-\left(-30\right)±\sqrt{900-4\times 7\times 27}}{2\times 7}$$

Multiply -4 times 7.

$$x=\frac{-\left(-30\right)±\sqrt{900-28\times 27}}{2\times 7}$$

Multiply -28 times 27.

$$x=\frac{-\left(-30\right)±\sqrt{900-756}}{2\times 7}$$

Add 900 to -756.

$$x=\frac{-\left(-30\right)±\sqrt{144}}{2\times 7}$$

Take the square root of 144.

$$x=\frac{-\left(-30\right)±12}{2\times 7}$$

The opposite of -30 is 30.

$$x=\frac{30±12}{2\times 7}$$

Multiply 2 times 7.

$$x=\frac{30±12}{14}$$

Now solve the equation x=\frac{30±12}{14} when ± is plus. Add 30 to 12.

$$x=\frac{42}{14}$$

Divide 42 by 14.

$$x=3$$

Now solve the equation x=\frac{30±12}{14} when ± is minus. Subtract 12 from 30.

$$x=\frac{18}{14}$$

Reduce the fraction \frac{18}{14} to lowest terms by extracting and canceling out 2.

$$x=\frac{9}{7}$$

The equation is now solved.

$$x=3 x=\frac{9}{7}$$

Steps for Completing the Square

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

$$3\left(x^{2}-6x+9\right)+4x\left(x-3\right)=0$$

Use the distributive property to multiply 3 by x^{2}-6x+9.

$$3x^{2}-18x+27+4x\left(x-3\right)=0$$

Use the distributive property to multiply 4x by x-3.

$$3x^{2}-18x+27+4x^{2}-12x=0$$

Combine 3x^{2} and 4x^{2} to get 7x^{2}.

$$7x^{2}-18x+27-12x=0$$

Combine -18x and -12x to get -30x.

$$7x^{2}-30x+27=0$$

Subtract 27 from both sides. Anything subtracted from zero gives its negation.

$$7x^{2}-30x=-27$$

Divide both sides by 7.

$$\frac{7x^{2}-30x}{7}=-\frac{27}{7}$$

Dividing by 7 undoes the multiplication by 7.

$$x^{2}-\frac{30}{7}x=-\frac{27}{7}$$

Divide -\frac{30}{7}, the coefficient of the x term, by 2 to get -\frac{15}{7}. Then add the square of -\frac{15}{7} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{30}{7}x+\left(-\frac{15}{7}\right)^{2}=-\frac{27}{7}+\left(-\frac{15}{7}\right)^{2}$$

Square -\frac{15}{7} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{30}{7}x+\frac{225}{49}=-\frac{27}{7}+\frac{225}{49}$$

Add -\frac{27}{7} to \frac{225}{49} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{30}{7}x+\frac{225}{49}=\frac{36}{49}$$

Factor x^{2}-\frac{30}{7}x+\frac{225}{49}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{15}{7}\right)^{2}=\frac{36}{49}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{15}{7}\right)^{2}}=\sqrt{\frac{36}{49}}$$

Simplify.

$$x-\frac{15}{7}=\frac{6}{7} x-\frac{15}{7}=-\frac{6}{7}$$

Add \frac{15}{7} to both sides of the equation.

$$x=3 x=\frac{9}{7}$$