Solve x-4[3(x-2)+11]=-x/3(x+5)

$$x - 4 [ 3 ( x - 2 ) + 11 ] = - \frac { x } { 3 } ( x + 5 )$$

Solve for x

$$x=-2$$

$$x=30$$

Steps Using Factoring

Multiply both sides of the equation by 3.

$$3x-12\left(3\left(x-2\right)+11\right)=3\left(-\frac{x}{3}\right)\left(x+5\right)$$

Use the distributive property to multiply 3 by x-2.

$$3x-12\left(3x-6+11\right)=3\left(-\frac{x}{3}\right)\left(x+5\right)$$

Add -6 and 11 to get 5.

$$3x-12\left(3x+5\right)=3\left(-\frac{x}{3}\right)\left(x+5\right)$$

Use the distributive property to multiply -12 by 3x+5.

$$3x-36x-60=3\left(-\frac{x}{3}\right)\left(x+5\right)$$

Combine 3x and -36x to get -33x.

$$-33x-60=3\left(-\frac{x}{3}\right)\left(x+5\right)$$

Express 3\left(-\frac{x}{3}\right) as a single fraction.

$$-33x-60=\frac{-3x}{3}\left(x+5\right)$$

Cancel out 3 and 3.

$$-33x-60=-x\left(x+5\right)$$

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

$$-33x-60=-x^{2}-5x$$

Add x^{2} to both sides.

$$-33x-60+x^{2}=-5x$$

Add 5x to both sides.

$$-33x-60+x^{2}+5x=0$$

Combine -33x and 5x to get -28x.

$$-28x-60+x^{2}=0$$

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

$$x^{2}-28x-60=0$$

To solve the equation, factor x^{2}-28x-60 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-28 ab=-60$$

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 -60.

$$1,-60 2,-30 3,-20 4,-15 5,-12 6,-10$$

Calculate the sum for each pair.

$$1-60=-59 2-30=-28 3-20=-17 4-15=-11 5-12=-7 6-10=-4$$

The solution is the pair that gives sum -28.

$$a=-30 b=2$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-30\right)\left(x+2\right)$$

To find equation solutions, solve x-30=0 and x+2=0.

$$x=30 x=-2$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 3.

$$3x-12\left(3\left(x-2\right)+11\right)=3\left(-\frac{x}{3}\right)\left(x+5\right)$$

Use the distributive property to multiply 3 by x-2.

$$3x-12\left(3x-6+11\right)=3\left(-\frac{x}{3}\right)\left(x+5\right)$$

Add -6 and 11 to get 5.

$$3x-12\left(3x+5\right)=3\left(-\frac{x}{3}\right)\left(x+5\right)$$

Use the distributive property to multiply -12 by 3x+5.

$$3x-36x-60=3\left(-\frac{x}{3}\right)\left(x+5\right)$$

Combine 3x and -36x to get -33x.

$$-33x-60=3\left(-\frac{x}{3}\right)\left(x+5\right)$$

Express 3\left(-\frac{x}{3}\right) as a single fraction.

$$-33x-60=\frac{-3x}{3}\left(x+5\right)$$

Cancel out 3 and 3.

$$-33x-60=-x\left(x+5\right)$$

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

$$-33x-60=-x^{2}-5x$$

Add x^{2} to both sides.

$$-33x-60+x^{2}=-5x$$

Add 5x to both sides.

$$-33x-60+x^{2}+5x=0$$

Combine -33x and 5x to get -28x.

$$-28x-60+x^{2}=0$$

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

$$x^{2}-28x-60=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-60. To find a and b, set up a system to be solved.

$$a+b=-28 ab=1\left(-60\right)=-60$$

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 -60.

$$1,-60 2,-30 3,-20 4,-15 5,-12 6,-10$$

Calculate the sum for each pair.

$$1-60=-59 2-30=-28 3-20=-17 4-15=-11 5-12=-7 6-10=-4$$

The solution is the pair that gives sum -28.

$$a=-30 b=2$$

Rewrite x^{2}-28x-60 as \left(x^{2}-30x\right)+\left(2x-60\right).

$$\left(x^{2}-30x\right)+\left(2x-60\right)$$

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

$$x\left(x-30\right)+2\left(x-30\right)$$

Factor out common term x-30 by using distributive property.

$$\left(x-30\right)\left(x+2\right)$$

To find equation solutions, solve x-30=0 and x+2=0.

$$x=30 x=-2$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 3.

$$3x-12\left(3\left(x-2\right)+11\right)=3\left(-\frac{x}{3}\right)\left(x+5\right)$$

Use the distributive property to multiply 3 by x-2.

$$3x-12\left(3x-6+11\right)=3\left(-\frac{x}{3}\right)\left(x+5\right)$$

Add -6 and 11 to get 5.

$$3x-12\left(3x+5\right)=3\left(-\frac{x}{3}\right)\left(x+5\right)$$

Use the distributive property to multiply -12 by 3x+5.

$$3x-36x-60=3\left(-\frac{x}{3}\right)\left(x+5\right)$$

Combine 3x and -36x to get -33x.

$$-33x-60=3\left(-\frac{x}{3}\right)\left(x+5\right)$$

Express 3\left(-\frac{x}{3}\right) as a single fraction.

$$-33x-60=\frac{-3x}{3}\left(x+5\right)$$

Cancel out 3 and 3.

$$-33x-60=-x\left(x+5\right)$$

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

$$-33x-60=-x^{2}-5x$$

Add x^{2} to both sides.

$$-33x-60+x^{2}=-5x$$

Add 5x to both sides.

$$-33x-60+x^{2}+5x=0$$

Combine -33x and 5x to get -28x.

$$-28x-60+x^{2}=0$$

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.

$$x^{2}-28x-60=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -28 for b, and -60 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-28\right)±\sqrt{\left(-28\right)^{2}-4\left(-60\right)}}{2}$$

Square -28.

$$x=\frac{-\left(-28\right)±\sqrt{784-4\left(-60\right)}}{2}$$

Multiply -4 times -60.

$$x=\frac{-\left(-28\right)±\sqrt{784+240}}{2}$$

Add 784 to 240.

$$x=\frac{-\left(-28\right)±\sqrt{1024}}{2}$$

Take the square root of 1024.

$$x=\frac{-\left(-28\right)±32}{2}$$

The opposite of -28 is 28.

$$x=\frac{28±32}{2}$$

Now solve the equation x=\frac{28±32}{2} when ± is plus. Add 28 to 32.

$$x=\frac{60}{2}$$

Divide 60 by 2.

$$x=30$$

Now solve the equation x=\frac{28±32}{2} when ± is minus. Subtract 32 from 28.

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

Divide -4 by 2.

$$x=-2$$

The equation is now solved.

$$x=30 x=-2$$

Steps for Completing the Square

Multiply both sides of the equation by 3.

$$3x-12\left(3\left(x-2\right)+11\right)=3\left(-\frac{x}{3}\right)\left(x+5\right)$$

Use the distributive property to multiply 3 by x-2.

$$3x-12\left(3x-6+11\right)=3\left(-\frac{x}{3}\right)\left(x+5\right)$$

Add -6 and 11 to get 5.

$$3x-12\left(3x+5\right)=3\left(-\frac{x}{3}\right)\left(x+5\right)$$

Use the distributive property to multiply -12 by 3x+5.

$$3x-36x-60=3\left(-\frac{x}{3}\right)\left(x+5\right)$$

Combine 3x and -36x to get -33x.

$$-33x-60=3\left(-\frac{x}{3}\right)\left(x+5\right)$$

Express 3\left(-\frac{x}{3}\right) as a single fraction.

$$-33x-60=\frac{-3x}{3}\left(x+5\right)$$

Cancel out 3 and 3.

$$-33x-60=-x\left(x+5\right)$$

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

$$-33x-60=-x^{2}-5x$$

Add x^{2} to both sides.

$$-33x-60+x^{2}=-5x$$

Add 5x to both sides.

$$-33x-60+x^{2}+5x=0$$

Combine -33x and 5x to get -28x.

$$-28x-60+x^{2}=0$$

Add 60 to both sides. Anything plus zero gives itself.

$$-28x+x^{2}=60$$

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.

$$x^{2}-28x=60$$

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

$$x^{2}-28x+\left(-14\right)^{2}=60+\left(-14\right)^{2}$$

Square -14.

$$x^{2}-28x+196=60+196$$

Add 60 to 196.

$$x^{2}-28x+196=256$$

Factor x^{2}-28x+196. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-14\right)^{2}=256$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-14\right)^{2}}=\sqrt{256}$$

Simplify.

$$x-14=16 x-14=-16$$

Add 14 to both sides of the equation.

$$x=30 x=-2$$