Solve 25quad4(x+3)^2=44quad25

$$25 \quad 4 ( x + 3 ) ^ { 2 } = 44 \quad 25$$

Solve for x

$$x=\sqrt{11}-3\approx 0.31662479$$

$$x=-\left(\sqrt{11}+3\right)\approx -6.31662479$$

Steps Using the Quadratic Formula

Cancel out 25 on both sides.

$$4\left(x+3\right)^{2}=44$$

Divide both sides by 4.

$$\left(x+3\right)^{2}=\frac{44}{4}$$

Divide 44 by 4 to get 11.

$$\left(x+3\right)^{2}=11$$

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

$$x^{2}+6x+9=11$$

Subtract 11 from both sides.

$$x^{2}+6x+9-11=0$$

Subtract 11 from 9 to get -2.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 6 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-2\right)}}{2}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\left(-2\right)}}{2}$$

Multiply -4 times -2.

$$x=\frac{-6±\sqrt{36+8}}{2}$$

Add 36 to 8.

$$x=\frac{-6±\sqrt{44}}{2}$$

Take the square root of 44.

$$x=\frac{-6±2\sqrt{11}}{2}$$

Now solve the equation x=\frac{-6±2\sqrt{11}}{2} when ± is plus. Add -6 to 2\sqrt{11}.

$$x=\frac{2\sqrt{11}-6}{2}$$

Divide -6+2\sqrt{11} by 2.

$$x=\sqrt{11}-3$$

Now solve the equation x=\frac{-6±2\sqrt{11}}{2} when ± is minus. Subtract 2\sqrt{11} from -6.

$$x=\frac{-2\sqrt{11}-6}{2}$$

Divide -6-2\sqrt{11} by 2.

$$x=-\sqrt{11}-3$$

The equation is now solved.

$$x=\sqrt{11}-3 x=-\sqrt{11}-3$$

Steps for Completing the Square

Cancel out 25 on both sides.

$$4\left(x+3\right)^{2}=44$$

Divide both sides by 4.

$$\left(x+3\right)^{2}=\frac{44}{4}$$

Divide 44 by 4 to get 11.

$$\left(x+3\right)^{2}=11$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+3\right)^{2}}=\sqrt{11}$$

Simplify.

$$x+3=\sqrt{11} x+3=-\sqrt{11}$$

Subtract 3 from both sides of the equation.

$$x=\sqrt{11}-3 x=-\sqrt{11}-3$$