Solve (2+m)^2-4(+m+3)=0

$$( 2 + m ) ^ { 2 } - 4 ( + m + 3 ) = 0$$

Solve for m

$$m=2\sqrt{2}\approx 2.828427125$$

$$m=-2\sqrt{2}\approx -2.828427125$$

Steps by Finding Square Root

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

$$4+4m+m^{2}-4\left(m+3\right)=0$$

Use the distributive property to multiply -4 by m+3.

$$4+4m+m^{2}-4m-12=0$$

Combine 4m and -4m to get 0.

$$4+m^{2}-12=0$$

Subtract 12 from 4 to get -8.

$$-8+m^{2}=0$$

Add 8 to both sides. Anything plus zero gives itself.

$$m^{2}=8$$

Take the square root of both sides of the equation.

$$m=2\sqrt{2} m=-2\sqrt{2}$$

Steps Using the Quadratic Formula

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

$$4+4m+m^{2}-4\left(m+3\right)=0$$

Use the distributive property to multiply -4 by m+3.

$$4+4m+m^{2}-4m-12=0$$

Combine 4m and -4m to get 0.

$$4+m^{2}-12=0$$

Subtract 12 from 4 to get -8.

$$-8+m^{2}=0$$

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

$$m^{2}-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$m=\frac{0±\sqrt{0^{2}-4\left(-8\right)}}{2}$$

Square 0.

$$m=\frac{0±\sqrt{-4\left(-8\right)}}{2}$$

Multiply -4 times -8.

$$m=\frac{0±\sqrt{32}}{2}$$

Take the square root of 32.

$$m=\frac{0±4\sqrt{2}}{2}$$

Now solve the equation m=\frac{0±4\sqrt{2}}{2} when ± is plus.

$$m=2\sqrt{2}$$

Now solve the equation m=\frac{0±4\sqrt{2}}{2} when ± is minus.

$$m=-2\sqrt{2}$$

The equation is now solved.

$$m=2\sqrt{2} m=-2\sqrt{2}$$