Solve (2x-5)(x+9)+4x^2=(2x-5)^2+25

$$( 2 x - 5 ) ( x + 9 ) + 4 x ^ { 2 } = ( 2 x - 5 ) ^ { 2 } + 25$$

Solve for x

$$x=-19$$

$$x = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

Steps Using Factoring By Grouping

Use the distributive property to multiply 2x-5 by x+9 and combine like terms.

$$2x^{2}+13x-45+4x^{2}=\left(2x-5\right)^{2}+25$$

Combine 2x^{2} and 4x^{2} to get 6x^{2}.

$$6x^{2}+13x-45=\left(2x-5\right)^{2}+25$$

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

$$6x^{2}+13x-45=4x^{2}-20x+25+25$$

Add 25 and 25 to get 50.

$$6x^{2}+13x-45=4x^{2}-20x+50$$

Subtract 4x^{2} from both sides.

$$6x^{2}+13x-45-4x^{2}=-20x+50$$

Combine 6x^{2} and -4x^{2} to get 2x^{2}.

$$2x^{2}+13x-45=-20x+50$$

Add 20x to both sides.

$$2x^{2}+13x-45+20x=50$$

Combine 13x and 20x to get 33x.

$$2x^{2}+33x-45=50$$

Subtract 50 from both sides.

$$2x^{2}+33x-45-50=0$$

Subtract 50 from -45 to get -95.

$$2x^{2}+33x-95=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-95. To find a and b, set up a system to be solved.

$$a+b=33 ab=2\left(-95\right)=-190$$

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. List all such integer pairs that give product -190.

$$-1,190 -2,95 -5,38 -10,19$$

Calculate the sum for each pair.

$$-1+190=189 -2+95=93 -5+38=33 -10+19=9$$

The solution is the pair that gives sum 33.

$$a=-5 b=38$$

Rewrite 2x^{2}+33x-95 as \left(2x^{2}-5x\right)+\left(38x-95\right).

$$\left(2x^{2}-5x\right)+\left(38x-95\right)$$

Factor out x in the first and 19 in the second group.

$$x\left(2x-5\right)+19\left(2x-5\right)$$

Factor out common term 2x-5 by using distributive property.

$$\left(2x-5\right)\left(x+19\right)$$

To find equation solutions, solve 2x-5=0 and x+19=0.

$$x=\frac{5}{2} x=-19$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2x-5 by x+9 and combine like terms.

$$2x^{2}+13x-45+4x^{2}=\left(2x-5\right)^{2}+25$$

Combine 2x^{2} and 4x^{2} to get 6x^{2}.

$$6x^{2}+13x-45=\left(2x-5\right)^{2}+25$$

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

$$6x^{2}+13x-45=4x^{2}-20x+25+25$$

Add 25 and 25 to get 50.

$$6x^{2}+13x-45=4x^{2}-20x+50$$

Subtract 4x^{2} from both sides.

$$6x^{2}+13x-45-4x^{2}=-20x+50$$

Combine 6x^{2} and -4x^{2} to get 2x^{2}.

$$2x^{2}+13x-45=-20x+50$$

Add 20x to both sides.

$$2x^{2}+13x-45+20x=50$$

Combine 13x and 20x to get 33x.

$$2x^{2}+33x-45=50$$

Subtract 50 from both sides.

$$2x^{2}+33x-45-50=0$$

Subtract 50 from -45 to get -95.

$$2x^{2}+33x-95=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 33 for b, and -95 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-33±\sqrt{33^{2}-4\times 2\left(-95\right)}}{2\times 2}$$

Square 33.

$$x=\frac{-33±\sqrt{1089-4\times 2\left(-95\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-33±\sqrt{1089-8\left(-95\right)}}{2\times 2}$$

Multiply -8 times -95.

$$x=\frac{-33±\sqrt{1089+760}}{2\times 2}$$

Add 1089 to 760.

$$x=\frac{-33±\sqrt{1849}}{2\times 2}$$

Take the square root of 1849.

$$x=\frac{-33±43}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-33±43}{4}$$

Now solve the equation x=\frac{-33±43}{4} when ± is plus. Add -33 to 43.

$$x=\frac{10}{4}$$

Reduce the fraction \frac{10}{4} to lowest terms by extracting and canceling out 2.

$$x=\frac{5}{2}$$

Now solve the equation x=\frac{-33±43}{4} when ± is minus. Subtract 43 from -33.

$$x=-\frac{76}{4}$$

Divide -76 by 4.

$$x=-19$$

The equation is now solved.

$$x=\frac{5}{2} x=-19$$

Steps for Completing the Square

Use the distributive property to multiply 2x-5 by x+9 and combine like terms.

$$2x^{2}+13x-45+4x^{2}=\left(2x-5\right)^{2}+25$$

Combine 2x^{2} and 4x^{2} to get 6x^{2}.

$$6x^{2}+13x-45=\left(2x-5\right)^{2}+25$$

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

$$6x^{2}+13x-45=4x^{2}-20x+25+25$$

Add 25 and 25 to get 50.

$$6x^{2}+13x-45=4x^{2}-20x+50$$

Subtract 4x^{2} from both sides.

$$6x^{2}+13x-45-4x^{2}=-20x+50$$

Combine 6x^{2} and -4x^{2} to get 2x^{2}.

$$2x^{2}+13x-45=-20x+50$$

Add 20x to both sides.

$$2x^{2}+13x-45+20x=50$$

Combine 13x and 20x to get 33x.

$$2x^{2}+33x-45=50$$

Add 45 to both sides.

$$2x^{2}+33x=50+45$$

Add 50 and 45 to get 95.

$$2x^{2}+33x=95$$

Divide both sides by 2.

$$\frac{2x^{2}+33x}{2}=\frac{95}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{33}{2}x=\frac{95}{2}$$

Divide \frac{33}{2}, the coefficient of the x term, by 2 to get \frac{33}{4}. Then add the square of \frac{33}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{33}{2}x+\left(\frac{33}{4}\right)^{2}=\frac{95}{2}+\left(\frac{33}{4}\right)^{2}$$

Square \frac{33}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{33}{2}x+\frac{1089}{16}=\frac{95}{2}+\frac{1089}{16}$$

Add \frac{95}{2} to \frac{1089}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{33}{2}x+\frac{1089}{16}=\frac{1849}{16}$$

Factor x^{2}+\frac{33}{2}x+\frac{1089}{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}.

$$\left(x+\frac{33}{4}\right)^{2}=\frac{1849}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{33}{4}\right)^{2}}=\sqrt{\frac{1849}{16}}$$

Simplify.

$$x+\frac{33}{4}=\frac{43}{4} x+\frac{33}{4}=-\frac{43}{4}$$

Subtract \frac{33}{4} from both sides of the equation.

$$x=\frac{5}{2} x=-19$$