Solve 1/2[(x+1)^2+3^2]=(x+1)x

$$\frac { 1 } { 2 } [ ( x + 1 ) ^ { 2 } + 3 ^ { 2 } ] = ( x + 1 ) x$$

Solve for x

$$x=\sqrt{10}\approx 3.16227766$$

$$x=-\sqrt{10}\approx -3.16227766$$

Steps by Finding Square Root

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

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

Calculate 3 to the power of 2 and get 9.

$$\frac{1}{2}\left(x^{2}+2x+1+9\right)=\left(x+1\right)x$$

Add 1 and 9 to get 10.

$$\frac{1}{2}\left(x^{2}+2x+10\right)=\left(x+1\right)x$$

Use the distributive property to multiply \frac{1}{2} by x^{2}+2x+10.

$$\frac{1}{2}x^{2}+x+5=\left(x+1\right)x$$

Use the distributive property to multiply x+1 by x.

$$\frac{1}{2}x^{2}+x+5=x^{2}+x$$

Subtract x^{2} from both sides.

$$\frac{1}{2}x^{2}+x+5-x^{2}=x$$

Combine \frac{1}{2}x^{2} and -x^{2} to get -\frac{1}{2}x^{2}.

$$-\frac{1}{2}x^{2}+x+5=x$$

Subtract x from both sides.

$$-\frac{1}{2}x^{2}+x+5-x=0$$

Combine x and -x to get 0.

$$-\frac{1}{2}x^{2}+5=0$$

Subtract 5 from both sides. Anything subtracted from zero gives its negation.

$$-\frac{1}{2}x^{2}=-5$$

Multiply both sides by -2, the reciprocal of -\frac{1}{2}.

$$x^{2}=-5\left(-2\right)$$

Multiply -5 and -2 to get 10.

$$x^{2}=10$$

Take the square root of both sides of the equation.

$$x=\sqrt{10} x=-\sqrt{10}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

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

Calculate 3 to the power of 2 and get 9.

$$\frac{1}{2}\left(x^{2}+2x+1+9\right)=\left(x+1\right)x$$

Add 1 and 9 to get 10.

$$\frac{1}{2}\left(x^{2}+2x+10\right)=\left(x+1\right)x$$

Use the distributive property to multiply \frac{1}{2} by x^{2}+2x+10.

$$\frac{1}{2}x^{2}+x+5=\left(x+1\right)x$$

Use the distributive property to multiply x+1 by x.

$$\frac{1}{2}x^{2}+x+5=x^{2}+x$$

Subtract x^{2} from both sides.

$$\frac{1}{2}x^{2}+x+5-x^{2}=x$$

Combine \frac{1}{2}x^{2} and -x^{2} to get -\frac{1}{2}x^{2}.

$$-\frac{1}{2}x^{2}+x+5=x$$

Subtract x from both sides.

$$-\frac{1}{2}x^{2}+x+5-x=0$$

Combine x and -x to get 0.

$$-\frac{1}{2}x^{2}+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{1}{2} for a, 0 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-\frac{1}{2}\right)\times 5}}{2\left(-\frac{1}{2}\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-\frac{1}{2}\right)\times 5}}{2\left(-\frac{1}{2}\right)}$$

Multiply -4 times -\frac{1}{2}.

$$x=\frac{0±\sqrt{2\times 5}}{2\left(-\frac{1}{2}\right)}$$

Multiply 2 times 5.

$$x=\frac{0±\sqrt{10}}{2\left(-\frac{1}{2}\right)}$$

Multiply 2 times -\frac{1}{2}.

$$x=\frac{0±\sqrt{10}}{-1}$$

Now solve the equation x=\frac{0±\sqrt{10}}{-1} when ± is plus.

$$x=-\sqrt{10}$$

Now solve the equation x=\frac{0±\sqrt{10}}{-1} when ± is minus.

$$x=\sqrt{10}$$

The equation is now solved.

$$x=-\sqrt{10} x=\sqrt{10}$$