Solve (x+4)(2x-1)=(-x-7)(4+x

$$( x + 4 ) ( 2 x - 1 ) = ( - x - 7 ) ( 4 + x$$

Solve for x

$$x=-4$$

$$x=-2$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x+4 by 2x-1 and combine like terms.

$$2x^{2}+7x-4=\left(-x-7\right)\left(4+x\right)$$

Use the distributive property to multiply -x-7 by 4+x.

$$2x^{2}+7x-4=4\left(-x\right)+\left(-x\right)x-28-7x$$

Subtract 4\left(-x\right) from both sides.

$$2x^{2}+7x-4-4\left(-x\right)=\left(-x\right)x-28-7x$$

Subtract \left(-x\right)x from both sides.

$$2x^{2}+7x-4-4\left(-x\right)-\left(-x\right)x=-28-7x$$

Subtract -28 from both sides.

$$2x^{2}+7x-4-4\left(-x\right)-\left(-x\right)x-\left(-28\right)=-7x$$

The opposite of -28 is 28.

$$2x^{2}+7x-4-4\left(-x\right)-\left(-x\right)x+28=-7x$$

Add 7x to both sides.

$$2x^{2}+7x-4-4\left(-x\right)-\left(-x\right)x+28+7x=0$$

Multiply -1 and 4 to get -4.

$$2x^{2}+7x-4-4\left(-1\right)x-\left(-xx\right)+28+7x=0$$

Multiply -4 and -1 to get 4.

$$2x^{2}+7x-4+4x-\left(-xx\right)+28+7x=0$$

Combine 7x and 4x to get 11x.

$$2x^{2}+11x-4-\left(-xx\right)+28+7x=0$$

Multiply x and x to get x^{2}.

$$2x^{2}+11x-4-\left(-x^{2}\right)+28+7x=0$$

Multiply -1 and -1 to get 1.

$$2x^{2}+11x-4+x^{2}+28+7x=0$$

Combine 2x^{2} and x^{2} to get 3x^{2}.

$$3x^{2}+11x-4+28+7x=0$$

Add -4 and 28 to get 24.

$$3x^{2}+11x+24+7x=0$$

Combine 11x and 7x to get 18x.

$$3x^{2}+18x+24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 18 for b, and 24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-18±\sqrt{18^{2}-4\times 3\times 24}}{2\times 3}$$

Square 18.

$$x=\frac{-18±\sqrt{324-4\times 3\times 24}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-18±\sqrt{324-12\times 24}}{2\times 3}$$

Multiply -12 times 24.

$$x=\frac{-18±\sqrt{324-288}}{2\times 3}$$

Add 324 to -288.

$$x=\frac{-18±\sqrt{36}}{2\times 3}$$

Take the square root of 36.

$$x=\frac{-18±6}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-18±6}{6}$$

Now solve the equation x=\frac{-18±6}{6} when ± is plus. Add -18 to 6.

$$x=-\frac{12}{6}$$

Divide -12 by 6.

$$x=-2$$

Now solve the equation x=\frac{-18±6}{6} when ± is minus. Subtract 6 from -18.

$$x=-\frac{24}{6}$$

Divide -24 by 6.

$$x=-4$$

The equation is now solved.

$$x=-2 x=-4$$

Steps for Completing the Square

Use the distributive property to multiply x+4 by 2x-1 and combine like terms.

$$2x^{2}+7x-4=\left(-x-7\right)\left(4+x\right)$$

Use the distributive property to multiply -x-7 by 4+x.

$$2x^{2}+7x-4=4\left(-x\right)+\left(-x\right)x-28-7x$$

Subtract 4\left(-x\right) from both sides.

$$2x^{2}+7x-4-4\left(-x\right)=\left(-x\right)x-28-7x$$

Subtract \left(-x\right)x from both sides.

$$2x^{2}+7x-4-4\left(-x\right)-\left(-x\right)x=-28-7x$$

Add 7x to both sides.

$$2x^{2}+7x-4-4\left(-x\right)-\left(-x\right)x+7x=-28$$

Multiply -1 and 4 to get -4.

$$2x^{2}+7x-4-4\left(-1\right)x-\left(-xx\right)+7x=-28$$

Multiply -4 and -1 to get 4.

$$2x^{2}+7x-4+4x-\left(-xx\right)+7x=-28$$

Combine 7x and 4x to get 11x.

$$2x^{2}+11x-4-\left(-xx\right)+7x=-28$$

Multiply x and x to get x^{2}.

$$2x^{2}+11x-4-\left(-x^{2}\right)+7x=-28$$

Multiply -1 and -1 to get 1.

$$2x^{2}+11x-4+x^{2}+7x=-28$$

Combine 2x^{2} and x^{2} to get 3x^{2}.

$$3x^{2}+11x-4+7x=-28$$

Combine 11x and 7x to get 18x.

$$3x^{2}+18x-4=-28$$

Add 4 to both sides.

$$3x^{2}+18x=-28+4$$

Add -28 and 4 to get -24.

$$3x^{2}+18x=-24$$

Divide both sides by 3.

$$\frac{3x^{2}+18x}{3}=-\frac{24}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{18}{3}x=-\frac{24}{3}$$

Divide 18 by 3.

$$x^{2}+6x=-\frac{24}{3}$$

Divide -24 by 3.

$$x^{2}+6x=-8$$

Divide 6, the coefficient of the x term, by 2 to get 3. Then add the square of 3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+6x+3^{2}=-8+3^{2}$$

Square 3.

$$x^{2}+6x+9=-8+9$$

Add -8 to 9.

$$x^{2}+6x+9=1$$

Factor x^{2}+6x+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x+3=1 x+3=-1$$

Subtract 3 from both sides of the equation.

$$x=-2 x=-4$$