Solve left(2x-1right)^2-left(x-2right)left(5x+2right)=0

$${ \left( 2x-1 \right) }^{ 2 } - \left( x-2 \right) \left( 5x+2 \right) = 0$$

Solve for x

$$x=-1$$

$$x=5$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-1\right)^{2}.

$$4x^{2}-4x+1-\left(x-2\right)\left(5x+2\right)=0$$

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

$$4x^{2}-4x+1-\left(5x^{2}-8x-4\right)=0$$

To find the opposite of 5x^{2}-8x-4, find the opposite of each term.

$$4x^{2}-4x+1-5x^{2}+8x+4=0$$

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

$$-x^{2}-4x+1+8x+4=0$$

Combine -4x and 8x to get 4x.

$$-x^{2}+4x+1+4=0$$

Add 1 and 4 to get 5.

$$-x^{2}+4x+5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+5. To find a and b, set up a system to be solved.

$$a+b=4 ab=-5=-5$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=5 b=-1$$

Rewrite -x^{2}+4x+5 as \left(-x^{2}+5x\right)+\left(-x+5\right).

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

Factor out -x in the first and -1 in the second group.

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

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

$$\left(x-5\right)\left(-x-1\right)$$

To find equation solutions, solve x-5=0 and -x-1=0.

$$x=5 x=-1$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-1\right)^{2}.

$$4x^{2}-4x+1-\left(x-2\right)\left(5x+2\right)=0$$

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

$$4x^{2}-4x+1-\left(5x^{2}-8x-4\right)=0$$

To find the opposite of 5x^{2}-8x-4, find the opposite of each term.

$$4x^{2}-4x+1-5x^{2}+8x+4=0$$

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

$$-x^{2}-4x+1+8x+4=0$$

Combine -4x and 8x to get 4x.

$$-x^{2}+4x+1+4=0$$

Add 1 and 4 to get 5.

$$-x^{2}+4x+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 4 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\left(-1\right)\times 5}}{2\left(-1\right)}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\left(-1\right)\times 5}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-4±\sqrt{16+4\times 5}}{2\left(-1\right)}$$

Multiply 4 times 5.

$$x=\frac{-4±\sqrt{16+20}}{2\left(-1\right)}$$

Add 16 to 20.

$$x=\frac{-4±\sqrt{36}}{2\left(-1\right)}$$

Take the square root of 36.

$$x=\frac{-4±6}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-4±6}{-2}$$

Now solve the equation x=\frac{-4±6}{-2} when ± is plus. Add -4 to 6.

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

Divide 2 by -2.

$$x=-1$$

Now solve the equation x=\frac{-4±6}{-2} when ± is minus. Subtract 6 from -4.

$$x=-\frac{10}{-2}$$

Divide -10 by -2.

$$x=5$$

The equation is now solved.

$$x=-1 x=5$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-1\right)^{2}.

$$4x^{2}-4x+1-\left(x-2\right)\left(5x+2\right)=0$$

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

$$4x^{2}-4x+1-\left(5x^{2}-8x-4\right)=0$$

To find the opposite of 5x^{2}-8x-4, find the opposite of each term.

$$4x^{2}-4x+1-5x^{2}+8x+4=0$$

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

$$-x^{2}-4x+1+8x+4=0$$

Combine -4x and 8x to get 4x.

$$-x^{2}+4x+1+4=0$$

Add 1 and 4 to get 5.

$$-x^{2}+4x+5=0$$

Subtract 5 from both sides. Anything subtracted from zero gives its negation.

$$-x^{2}+4x=-5$$

Divide both sides by -1.

$$\frac{-x^{2}+4x}{-1}=-\frac{5}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{4}{-1}x=-\frac{5}{-1}$$

Divide 4 by -1.

$$x^{2}-4x=-\frac{5}{-1}$$

Divide -5 by -1.

$$x^{2}-4x=5$$

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

$$x^{2}-4x+\left(-2\right)^{2}=5+\left(-2\right)^{2}$$

Square -2.

$$x^{2}-4x+4=5+4$$

Add 5 to 4.

$$x^{2}-4x+4=9$$

Factor x^{2}-4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-2\right)^{2}=9$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{9}$$

Simplify.

$$x-2=3 x-2=-3$$

Add 2 to both sides of the equation.

$$x=5 x=-1$$