Solve 2(x+14)-(2x-1)(x+2)=2(x-1)

$$2 ( x + 14 ) - ( 2 x - 1 ) ( x + 2 ) = 2 ( x - 1 )$$

Solve for x

$$x = \frac{\sqrt{265} - 3}{4} \approx 3.319705149$$

$$x=\frac{-\sqrt{265}-3}{4}\approx -4.819705149$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2 by x+14.

$$2x+28-\left(2x-1\right)\left(x+2\right)=2\left(x-1\right)$$

Use the distributive property to multiply 2x-1 by x+2 and combine like terms.

$$2x+28-\left(2x^{2}+3x-2\right)=2\left(x-1\right)$$

To find the opposite of 2x^{2}+3x-2, find the opposite of each term.

$$2x+28-2x^{2}-3x+2=2\left(x-1\right)$$

Combine 2x and -3x to get -x.

$$-x+28-2x^{2}+2=2\left(x-1\right)$$

Add 28 and 2 to get 30.

$$-x+30-2x^{2}=2\left(x-1\right)$$

Use the distributive property to multiply 2 by x-1.

$$-x+30-2x^{2}=2x-2$$

Subtract 2x from both sides.

$$-x+30-2x^{2}-2x=-2$$

Combine -x and -2x to get -3x.

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

Add 2 to both sides.

$$-3x+30-2x^{2}+2=0$$

Add 30 and 2 to get 32.

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

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.

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

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

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

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-2\right)\times 32}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-\left(-3\right)±\sqrt{9+8\times 32}}{2\left(-2\right)}$$

Multiply 8 times 32.

$$x=\frac{-\left(-3\right)±\sqrt{9+256}}{2\left(-2\right)}$$

Add 9 to 256.

$$x=\frac{-\left(-3\right)±\sqrt{265}}{2\left(-2\right)}$$

The opposite of -3 is 3.

$$x=\frac{3±\sqrt{265}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{3±\sqrt{265}}{-4}$$

Now solve the equation x=\frac{3±\sqrt{265}}{-4} when ± is plus. Add 3 to \sqrt{265}.

$$x=\frac{\sqrt{265}+3}{-4}$$

Divide 3+\sqrt{265} by -4.

$$x=\frac{-\sqrt{265}-3}{4}$$

Now solve the equation x=\frac{3±\sqrt{265}}{-4} when ± is minus. Subtract \sqrt{265} from 3.

$$x=\frac{3-\sqrt{265}}{-4}$$

Divide 3-\sqrt{265} by -4.

$$x=\frac{\sqrt{265}-3}{4}$$

The equation is now solved.

$$x=\frac{-\sqrt{265}-3}{4} x=\frac{\sqrt{265}-3}{4}$$

Steps for Completing the Square

Use the distributive property to multiply 2 by x+14.

$$2x+28-\left(2x-1\right)\left(x+2\right)=2\left(x-1\right)$$

Use the distributive property to multiply 2x-1 by x+2 and combine like terms.

$$2x+28-\left(2x^{2}+3x-2\right)=2\left(x-1\right)$$

To find the opposite of 2x^{2}+3x-2, find the opposite of each term.

$$2x+28-2x^{2}-3x+2=2\left(x-1\right)$$

Combine 2x and -3x to get -x.

$$-x+28-2x^{2}+2=2\left(x-1\right)$$

Add 28 and 2 to get 30.

$$-x+30-2x^{2}=2\left(x-1\right)$$

Use the distributive property to multiply 2 by x-1.

$$-x+30-2x^{2}=2x-2$$

Subtract 2x from both sides.

$$-x+30-2x^{2}-2x=-2$$

Combine -x and -2x to get -3x.

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

Subtract 30 from both sides.

$$-3x-2x^{2}=-2-30$$

Subtract 30 from -2 to get -32.

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

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.

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

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

Divide -3 by -2.

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

Divide -32 by -2.

$$x^{2}+\frac{3}{2}x=16$$

Divide \frac{3}{2}, the coefficient of the x term, by 2 to get \frac{3}{4}. Then add the square of \frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{3}{2}x+\left(\frac{3}{4}\right)^{2}=16+\left(\frac{3}{4}\right)^{2}$$

Square \frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=16+\frac{9}{16}$$

Add 16 to \frac{9}{16}.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=\frac{265}{16}$$

Factor x^{2}+\frac{3}{2}x+\frac{9}{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}.

$$\left(x+\frac{3}{4}\right)^{2}=\frac{265}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{4}\right)^{2}}=\sqrt{\frac{265}{16}}$$

Simplify.

$$x+\frac{3}{4}=\frac{\sqrt{265}}{4} x+\frac{3}{4}=-\frac{\sqrt{265}}{4}$$

Subtract \frac{3}{4} from both sides of the equation.

$$x=\frac{\sqrt{265}-3}{4} x=\frac{-\sqrt{265}-3}{4}$$