Solve {l}{(2x+2)(x+3)=96}{[4text{mark}}

$$\left. \begin{array}{l}{ ( 2 x + 2 ) ( x + 3 ) = 96 }\\{ [ 4 \text { mark } }\end{array} \right.$$

Solve for x

$$x=-9$$

$$x=5$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2x+2 by x+3 and combine like terms.

$$2x^{2}+8x+6=96$$

Subtract 96 from both sides.

$$2x^{2}+8x+6-96=0$$

Subtract 96 from 6 to get -90.

$$2x^{2}+8x-90=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 8 for b, and -90 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\times 2\left(-90\right)}}{2\times 2}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\times 2\left(-90\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-8±\sqrt{64-8\left(-90\right)}}{2\times 2}$$

Multiply -8 times -90.

$$x=\frac{-8±\sqrt{64+720}}{2\times 2}$$

Add 64 to 720.

$$x=\frac{-8±\sqrt{784}}{2\times 2}$$

Take the square root of 784.

$$x=\frac{-8±28}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-8±28}{4}$$

Now solve the equation x=\frac{-8±28}{4} when ± is plus. Add -8 to 28.

$$x=\frac{20}{4}$$

Divide 20 by 4.

$$x=5$$

Now solve the equation x=\frac{-8±28}{4} when ± is minus. Subtract 28 from -8.

$$x=-\frac{36}{4}$$

Divide -36 by 4.

$$x=-9$$

The equation is now solved.

$$x=5 x=-9$$

Steps for Completing the Square

Use the distributive property to multiply 2x+2 by x+3 and combine like terms.

$$2x^{2}+8x+6=96$$

Subtract 6 from both sides.

$$2x^{2}+8x=96-6$$

Subtract 6 from 96 to get 90.

$$2x^{2}+8x=90$$

Divide both sides by 2.

$$\frac{2x^{2}+8x}{2}=\frac{90}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{8}{2}x=\frac{90}{2}$$

Divide 8 by 2.

$$x^{2}+4x=\frac{90}{2}$$

Divide 90 by 2.

$$x^{2}+4x=45$$

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.

$$x^{2}+4x+2^{2}=45+2^{2}$$

Square 2.

$$x^{2}+4x+4=45+4$$

Add 45 to 4.

$$x^{2}+4x+4=49$$

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}.

$$\left(x+2\right)^{2}=49$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{49}$$

Simplify.

$$x+2=7 x+2=-7$$

Subtract 2 from both sides of the equation.

$$x=5 x=-9$$