Solve (x-2)^2+2x(x+2)=2*(3x+10)

$$( x - 2 ) ^ { 2 } + 2 x ( x + 2 ) = 2 \cdot ( 3 x + 10 )$$

Solve for x

$$x=\frac{\sqrt{57}}{3}+1\approx 3.516611478$$

$$x=-\frac{\sqrt{57}}{3}+1\approx -1.516611478$$

Steps Using the Quadratic Formula

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

$$x^{2}-4x+4+2x\left(x+2\right)=2\left(3x+10\right)$$

Use the distributive property to multiply 2x by x+2.

$$x^{2}-4x+4+2x^{2}+4x=2\left(3x+10\right)$$

Combine x^{2} and 2x^{2} to get 3x^{2}.

$$3x^{2}-4x+4+4x=2\left(3x+10\right)$$

Combine -4x and 4x to get 0.

$$3x^{2}+4=2\left(3x+10\right)$$

Use the distributive property to multiply 2 by 3x+10.

$$3x^{2}+4=6x+20$$

Subtract 6x from both sides.

$$3x^{2}+4-6x=20$$

Subtract 20 from both sides.

$$3x^{2}+4-6x-20=0$$

Subtract 20 from 4 to get -16.

$$3x^{2}-16-6x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$3x^{2}-6x-16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -6 for b, and -16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 3\left(-16\right)}}{2\times 3}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 3\left(-16\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-6\right)±\sqrt{36-12\left(-16\right)}}{2\times 3}$$

Multiply -12 times -16.

$$x=\frac{-\left(-6\right)±\sqrt{36+192}}{2\times 3}$$

Add 36 to 192.

$$x=\frac{-\left(-6\right)±\sqrt{228}}{2\times 3}$$

Take the square root of 228.

$$x=\frac{-\left(-6\right)±2\sqrt{57}}{2\times 3}$$

The opposite of -6 is 6.

$$x=\frac{6±2\sqrt{57}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{6±2\sqrt{57}}{6}$$

Now solve the equation x=\frac{6±2\sqrt{57}}{6} when ± is plus. Add 6 to 2\sqrt{57}.

$$x=\frac{2\sqrt{57}+6}{6}$$

Divide 6+2\sqrt{57} by 6.

$$x=\frac{\sqrt{57}}{3}+1$$

Now solve the equation x=\frac{6±2\sqrt{57}}{6} when ± is minus. Subtract 2\sqrt{57} from 6.

$$x=\frac{6-2\sqrt{57}}{6}$$

Divide 6-2\sqrt{57} by 6.

$$x=-\frac{\sqrt{57}}{3}+1$$

The equation is now solved.

$$x=\frac{\sqrt{57}}{3}+1 x=-\frac{\sqrt{57}}{3}+1$$

Steps for Completing the Square

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

$$x^{2}-4x+4+2x\left(x+2\right)=2\left(3x+10\right)$$

Use the distributive property to multiply 2x by x+2.

$$x^{2}-4x+4+2x^{2}+4x=2\left(3x+10\right)$$

Combine x^{2} and 2x^{2} to get 3x^{2}.

$$3x^{2}-4x+4+4x=2\left(3x+10\right)$$

Combine -4x and 4x to get 0.

$$3x^{2}+4=2\left(3x+10\right)$$

Use the distributive property to multiply 2 by 3x+10.

$$3x^{2}+4=6x+20$$

Subtract 6x from both sides.

$$3x^{2}+4-6x=20$$

Subtract 4 from both sides.

$$3x^{2}-6x=20-4$$

Subtract 4 from 20 to get 16.

$$3x^{2}-6x=16$$

Divide both sides by 3.

$$\frac{3x^{2}-6x}{3}=\frac{16}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\left(-\frac{6}{3}\right)x=\frac{16}{3}$$

Divide -6 by 3.

$$x^{2}-2x=\frac{16}{3}$$

Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-2x+1=\frac{16}{3}+1$$

Add \frac{16}{3} to 1.

$$x^{2}-2x+1=\frac{19}{3}$$

Factor x^{2}-2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-1\right)^{2}=\frac{19}{3}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{\frac{19}{3}}$$

Simplify.

$$x-1=\frac{\sqrt{57}}{3} x-1=-\frac{\sqrt{57}}{3}$$

Add 1 to both sides of the equation.

$$x=\frac{\sqrt{57}}{3}+1 x=-\frac{\sqrt{57}}{3}+1$$