Solve 3left(5-xright)^2=2(x-5)

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

Solve for x

$$x=5$$

$$x = \frac{17}{3} = 5\frac{2}{3} \approx 5.666666667$$

Steps Using Factoring By Grouping

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

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

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

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

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

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

Subtract 2x from both sides.

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

Combine -30x and -2x to get -32x.

$$75-32x+3x^{2}=-10$$

Add 10 to both sides.

$$75-32x+3x^{2}+10=0$$

Add 75 and 10 to get 85.

$$85-32x+3x^{2}=0$$

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

$$3x^{2}-32x+85=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx+85. To find a and b, set up a system to be solved.

$$a+b=-32 ab=3\times 85=255$$

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 255.

$$-1,-255 -3,-85 -5,-51 -15,-17$$

Calculate the sum for each pair.

$$-1-255=-256 -3-85=-88 -5-51=-56 -15-17=-32$$

The solution is the pair that gives sum -32.

$$a=-17 b=-15$$

Rewrite 3x^{2}-32x+85 as \left(3x^{2}-17x\right)+\left(-15x+85\right).

$$\left(3x^{2}-17x\right)+\left(-15x+85\right)$$

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

$$x\left(3x-17\right)-5\left(3x-17\right)$$

Factor out common term 3x-17 by using distributive property.

$$\left(3x-17\right)\left(x-5\right)$$

To find equation solutions, solve 3x-17=0 and x-5=0.

$$x=\frac{17}{3} x=5$$

Steps Using the Quadratic Formula

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

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

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

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

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

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

Subtract 2x from both sides.

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

Combine -30x and -2x to get -32x.

$$75-32x+3x^{2}=-10$$

Add 10 to both sides.

$$75-32x+3x^{2}+10=0$$

Add 75 and 10 to get 85.

$$85-32x+3x^{2}=0$$

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.

$$3x^{2}-32x+85=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -32 for b, and 85 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-32\right)±\sqrt{\left(-32\right)^{2}-4\times 3\times 85}}{2\times 3}$$

Square -32.

$$x=\frac{-\left(-32\right)±\sqrt{1024-4\times 3\times 85}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-32\right)±\sqrt{1024-12\times 85}}{2\times 3}$$

Multiply -12 times 85.

$$x=\frac{-\left(-32\right)±\sqrt{1024-1020}}{2\times 3}$$

Add 1024 to -1020.

$$x=\frac{-\left(-32\right)±\sqrt{4}}{2\times 3}$$

Take the square root of 4.

$$x=\frac{-\left(-32\right)±2}{2\times 3}$$

The opposite of -32 is 32.

$$x=\frac{32±2}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{32±2}{6}$$

Now solve the equation x=\frac{32±2}{6} when ± is plus. Add 32 to 2.

$$x=\frac{34}{6}$$

Reduce the fraction \frac{34}{6} to lowest terms by extracting and canceling out 2.

$$x=\frac{17}{3}$$

Now solve the equation x=\frac{32±2}{6} when ± is minus. Subtract 2 from 32.

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

Divide 30 by 6.

$$x=5$$

The equation is now solved.

$$x=\frac{17}{3} x=5$$

Steps for Completing the Square

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

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

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

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

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

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

Subtract 2x from both sides.

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

Combine -30x and -2x to get -32x.

$$75-32x+3x^{2}=-10$$

Subtract 75 from both sides.

$$-32x+3x^{2}=-10-75$$

Subtract 75 from -10 to get -85.

$$-32x+3x^{2}=-85$$

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.

$$3x^{2}-32x=-85$$

Divide both sides by 3.

$$\frac{3x^{2}-32x}{3}=-\frac{85}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{32}{3}x=-\frac{85}{3}$$

Divide -\frac{32}{3}, the coefficient of the x term, by 2 to get -\frac{16}{3}. Then add the square of -\frac{16}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{32}{3}x+\left(-\frac{16}{3}\right)^{2}=-\frac{85}{3}+\left(-\frac{16}{3}\right)^{2}$$

Square -\frac{16}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{32}{3}x+\frac{256}{9}=-\frac{85}{3}+\frac{256}{9}$$

Add -\frac{85}{3} to \frac{256}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{32}{3}x+\frac{256}{9}=\frac{1}{9}$$

Factor x^{2}-\frac{32}{3}x+\frac{256}{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}.

$$\left(x-\frac{16}{3}\right)^{2}=\frac{1}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{16}{3}\right)^{2}}=\sqrt{\frac{1}{9}}$$

Simplify.

$$x-\frac{16}{3}=\frac{1}{3} x-\frac{16}{3}=-\frac{1}{3}$$

Add \frac{16}{3} to both sides of the equation.

$$x=\frac{17}{3} x=5$$