Solve x^2=(15-2x)(2x-5)

$$x ^ { 2 } = ( 15 - 2 x ) ( 2 x - 5 )$$

Solve for x

$$x=3$$

$$x=5$$

Steps Using Factoring By Grouping

Use the distributive property to multiply 15-2x by 2x-5 and combine like terms.

$$x^{2}=40x-75-4x^{2}$$

Subtract 40x from both sides.

$$x^{2}-40x=-75-4x^{2}$$

Subtract -75 from both sides.

$$x^{2}-40x-\left(-75\right)=-4x^{2}$$

The opposite of -75 is 75.

$$x^{2}-40x+75=-4x^{2}$$

Add 4x^{2} to both sides.

$$x^{2}-40x+75+4x^{2}=0$$

Combine x^{2} and 4x^{2} to get 5x^{2}.

$$5x^{2}-40x+75=0$$

Divide both sides by 5.

$$x^{2}-8x+15=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+15. To find a and b, set up a system to be solved.

$$a+b=-8 ab=1\times 15=15$$

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 15.

$$-1,-15 -3,-5$$

Calculate the sum for each pair.

$$-1-15=-16 -3-5=-8$$

The solution is the pair that gives sum -8.

$$a=-5 b=-3$$

Rewrite x^{2}-8x+15 as \left(x^{2}-5x\right)+\left(-3x+15\right).

$$\left(x^{2}-5x\right)+\left(-3x+15\right)$$

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

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

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

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

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

$$x=5 x=3$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 15-2x by 2x-5 and combine like terms.

$$x^{2}=40x-75-4x^{2}$$

Subtract 40x from both sides.

$$x^{2}-40x=-75-4x^{2}$$

Subtract -75 from both sides.

$$x^{2}-40x-\left(-75\right)=-4x^{2}$$

The opposite of -75 is 75.

$$x^{2}-40x+75=-4x^{2}$$

Add 4x^{2} to both sides.

$$x^{2}-40x+75+4x^{2}=0$$

Combine x^{2} and 4x^{2} to get 5x^{2}.

$$5x^{2}-40x+75=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -40 for b, and 75 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-40\right)±\sqrt{\left(-40\right)^{2}-4\times 5\times 75}}{2\times 5}$$

Square -40.

$$x=\frac{-\left(-40\right)±\sqrt{1600-4\times 5\times 75}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-40\right)±\sqrt{1600-20\times 75}}{2\times 5}$$

Multiply -20 times 75.

$$x=\frac{-\left(-40\right)±\sqrt{1600-1500}}{2\times 5}$$

Add 1600 to -1500.

$$x=\frac{-\left(-40\right)±\sqrt{100}}{2\times 5}$$

Take the square root of 100.

$$x=\frac{-\left(-40\right)±10}{2\times 5}$$

The opposite of -40 is 40.

$$x=\frac{40±10}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{40±10}{10}$$

Now solve the equation x=\frac{40±10}{10} when ± is plus. Add 40 to 10.

$$x=\frac{50}{10}$$

Divide 50 by 10.

$$x=5$$

Now solve the equation x=\frac{40±10}{10} when ± is minus. Subtract 10 from 40.

$$x=\frac{30}{10}$$

Divide 30 by 10.

$$x=3$$

The equation is now solved.

$$x=5 x=3$$

Steps for Completing the Square

Use the distributive property to multiply 15-2x by 2x-5 and combine like terms.

$$x^{2}=40x-75-4x^{2}$$

Subtract 40x from both sides.

$$x^{2}-40x=-75-4x^{2}$$

Add 4x^{2} to both sides.

$$x^{2}-40x+4x^{2}=-75$$

Combine x^{2} and 4x^{2} to get 5x^{2}.

$$5x^{2}-40x=-75$$

Divide both sides by 5.

$$\frac{5x^{2}-40x}{5}=-\frac{75}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}+\left(-\frac{40}{5}\right)x=-\frac{75}{5}$$

Divide -40 by 5.

$$x^{2}-8x=-\frac{75}{5}$$

Divide -75 by 5.

$$x^{2}-8x=-15$$

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-8x+\left(-4\right)^{2}=-15+\left(-4\right)^{2}$$

Square -4.

$$x^{2}-8x+16=-15+16$$

Add -15 to 16.

$$x^{2}-8x+16=1$$

Factor x^{2}-8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-4\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-4\right)^{2}}=\sqrt{1}$$

Simplify.

$$x-4=1 x-4=-1$$

Add 4 to both sides of the equation.

$$x=5 x=3$$