Solve (2)(2)*(x+3)(x+3)=41

$$(2)(2) \times (x+3)(x+3)=41$$

Solve for x

$$x=\frac{\sqrt{41}}{2}-3\approx 0.201562119$$

$$x=-\frac{\sqrt{41}}{2}-3\approx -6.201562119$$

Steps Using the Quadratic Formula

Multiply x+3 and x+3 to get \left(x+3\right)^{2}.

$$2\times 2\left(x+3\right)^{2}=41$$

Multiply 2 and 2 to get 4.

$$4\left(x+3\right)^{2}=41$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$4\left(x^{2}+6x+9\right)=41$$

Use the distributive property to multiply 4 by x^{2}+6x+9.

$$4x^{2}+24x+36=41$$

Subtract 41 from both sides.

$$4x^{2}+24x+36-41=0$$

Subtract 41 from 36 to get -5.

$$4x^{2}+24x-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 24 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-24±\sqrt{24^{2}-4\times 4\left(-5\right)}}{2\times 4}$$

Square 24.

$$x=\frac{-24±\sqrt{576-4\times 4\left(-5\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-24±\sqrt{576-16\left(-5\right)}}{2\times 4}$$

Multiply -16 times -5.

$$x=\frac{-24±\sqrt{576+80}}{2\times 4}$$

Add 576 to 80.

$$x=\frac{-24±\sqrt{656}}{2\times 4}$$

Take the square root of 656.

$$x=\frac{-24±4\sqrt{41}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-24±4\sqrt{41}}{8}$$

Now solve the equation x=\frac{-24±4\sqrt{41}}{8} when ± is plus. Add -24 to 4\sqrt{41}.

$$x=\frac{4\sqrt{41}-24}{8}$$

Divide -24+4\sqrt{41} by 8.

$$x=\frac{\sqrt{41}}{2}-3$$

Now solve the equation x=\frac{-24±4\sqrt{41}}{8} when ± is minus. Subtract 4\sqrt{41} from -24.

$$x=\frac{-4\sqrt{41}-24}{8}$$

Divide -24-4\sqrt{41} by 8.

$$x=-\frac{\sqrt{41}}{2}-3$$

The equation is now solved.

$$x=\frac{\sqrt{41}}{2}-3 x=-\frac{\sqrt{41}}{2}-3$$

Steps for Completing the Square

Multiply x+3 and x+3 to get \left(x+3\right)^{2}.

$$2\times 2\left(x+3\right)^{2}=41$$

Multiply 2 and 2 to get 4.

$$4\left(x+3\right)^{2}=41$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$4\left(x^{2}+6x+9\right)=41$$

Use the distributive property to multiply 4 by x^{2}+6x+9.

$$4x^{2}+24x+36=41$$

Subtract 36 from both sides.

$$4x^{2}+24x=41-36$$

Subtract 36 from 41 to get 5.

$$4x^{2}+24x=5$$

Divide both sides by 4.

$$\frac{4x^{2}+24x}{4}=\frac{5}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{24}{4}x=\frac{5}{4}$$

Divide 24 by 4.

$$x^{2}+6x=\frac{5}{4}$$

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.

$$x^{2}+6x+3^{2}=\frac{5}{4}+3^{2}$$

Square 3.

$$x^{2}+6x+9=\frac{5}{4}+9$$

Add \frac{5}{4} to 9.

$$x^{2}+6x+9=\frac{41}{4}$$

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}.

$$\left(x+3\right)^{2}=\frac{41}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+3\right)^{2}}=\sqrt{\frac{41}{4}}$$

Simplify.

$$x+3=\frac{\sqrt{41}}{2} x+3=-\frac{\sqrt{41}}{2}$$

Subtract 3 from both sides of the equation.

$$x=\frac{\sqrt{41}}{2}-3 x=-\frac{\sqrt{41}}{2}-3$$