Solve 3(x-5)^2+2(x-5)=0

$$3 ( x - 5 ) ^ { 2 } + 2 ( x - 5 ) = 0$$

Solve for x

$$x = \frac{13}{3} = 4\frac{1}{3} \approx 4.333333333$$

$$x=5$$

Steps Using Factoring By Grouping

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

$$3\left(x^{2}-10x+25\right)+2\left(x-5\right)=0$$

Use the distributive property to multiply 3 by x^{2}-10x+25.

$$3x^{2}-30x+75+2\left(x-5\right)=0$$

Use the distributive property to multiply 2 by x-5.

$$3x^{2}-30x+75+2x-10=0$$

Combine -30x and 2x to get -28x.

$$3x^{2}-28x+75-10=0$$

Subtract 10 from 75 to get 65.

$$3x^{2}-28x+65=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx+65. To find a and b, set up a system to be solved.

$$a+b=-28 ab=3\times 65=195$$

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 195.

$$-1,-195 -3,-65 -5,-39 -13,-15$$

Calculate the sum for each pair.

$$-1-195=-196 -3-65=-68 -5-39=-44 -13-15=-28$$

The solution is the pair that gives sum -28.

$$a=-15 b=-13$$

Rewrite 3x^{2}-28x+65 as \left(3x^{2}-15x\right)+\left(-13x+65\right).

$$\left(3x^{2}-15x\right)+\left(-13x+65\right)$$

Factor out 3x in the first and -13 in the second group.

$$3x\left(x-5\right)-13\left(x-5\right)$$

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

$$\left(x-5\right)\left(3x-13\right)$$

To find equation solutions, solve x-5=0 and 3x-13=0.

$$x=5 x=\frac{13}{3}$$

Steps Using the Quadratic Formula

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

$$3\left(x^{2}-10x+25\right)+2\left(x-5\right)=0$$

Use the distributive property to multiply 3 by x^{2}-10x+25.

$$3x^{2}-30x+75+2\left(x-5\right)=0$$

Use the distributive property to multiply 2 by x-5.

$$3x^{2}-30x+75+2x-10=0$$

Combine -30x and 2x to get -28x.

$$3x^{2}-28x+75-10=0$$

Subtract 10 from 75 to get 65.

$$3x^{2}-28x+65=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -28 for b, and 65 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-28\right)±\sqrt{\left(-28\right)^{2}-4\times 3\times 65}}{2\times 3}$$

Square -28.

$$x=\frac{-\left(-28\right)±\sqrt{784-4\times 3\times 65}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-28\right)±\sqrt{784-12\times 65}}{2\times 3}$$

Multiply -12 times 65.

$$x=\frac{-\left(-28\right)±\sqrt{784-780}}{2\times 3}$$

Add 784 to -780.

$$x=\frac{-\left(-28\right)±\sqrt{4}}{2\times 3}$$

Take the square root of 4.

$$x=\frac{-\left(-28\right)±2}{2\times 3}$$

The opposite of -28 is 28.

$$x=\frac{28±2}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{28±2}{6}$$

Now solve the equation x=\frac{28±2}{6} when ± is plus. Add 28 to 2.

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

Divide 30 by 6.

$$x=5$$

Now solve the equation x=\frac{28±2}{6} when ± is minus. Subtract 2 from 28.

$$x=\frac{26}{6}$$

Reduce the fraction \frac{26}{6} to lowest terms by extracting and canceling out 2.

$$x=\frac{13}{3}$$

The equation is now solved.

$$x=5 x=\frac{13}{3}$$

Steps for Completing the Square

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

$$3\left(x^{2}-10x+25\right)+2\left(x-5\right)=0$$

Use the distributive property to multiply 3 by x^{2}-10x+25.

$$3x^{2}-30x+75+2\left(x-5\right)=0$$

Use the distributive property to multiply 2 by x-5.

$$3x^{2}-30x+75+2x-10=0$$

Combine -30x and 2x to get -28x.

$$3x^{2}-28x+75-10=0$$

Subtract 10 from 75 to get 65.

$$3x^{2}-28x+65=0$$

Subtract 65 from both sides. Anything subtracted from zero gives its negation.

$$3x^{2}-28x=-65$$

Divide both sides by 3.

$$\frac{3x^{2}-28x}{3}=-\frac{65}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{28}{3}x=-\frac{65}{3}$$

Divide -\frac{28}{3}, the coefficient of the x term, by 2 to get -\frac{14}{3}. Then add the square of -\frac{14}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{28}{3}x+\left(-\frac{14}{3}\right)^{2}=-\frac{65}{3}+\left(-\frac{14}{3}\right)^{2}$$

Square -\frac{14}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{28}{3}x+\frac{196}{9}=-\frac{65}{3}+\frac{196}{9}$$

Add -\frac{65}{3} to \frac{196}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{28}{3}x+\frac{196}{9}=\frac{1}{9}$$

Factor x^{2}-\frac{28}{3}x+\frac{196}{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}.

$$\left(x-\frac{14}{3}\right)^{2}=\frac{1}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{14}{3}\right)^{2}}=\sqrt{\frac{1}{9}}$$

Simplify.

$$x-\frac{14}{3}=\frac{1}{3} x-\frac{14}{3}=-\frac{1}{3}$$

Add \frac{14}{3} to both sides of the equation.

$$x=5 x=\frac{13}{3}$$