Solve left(2x+3right)^2=x^2+left(x+27-4xright)^2

$${ \left(2x+3 \right) }^{ 2 } = { x }^{ 2 } + { \left(x+27-4x \right) }^{ 2 }$$

Solve for x

$$x=5$$

$$x=24$$

Steps Using the Quadratic Formula

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

$$4x^{2}+12x+9=x^{2}+\left(x+27-4x\right)^{2}$$

Combine x and -4x to get -3x.

$$4x^{2}+12x+9=x^{2}+\left(-3x+27\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-3x+27\right)^{2}.

$$4x^{2}+12x+9=x^{2}+9x^{2}-162x+729$$

Combine x^{2} and 9x^{2} to get 10x^{2}.

$$4x^{2}+12x+9=10x^{2}-162x+729$$

Subtract 10x^{2} from both sides.

$$4x^{2}+12x+9-10x^{2}=-162x+729$$

Combine 4x^{2} and -10x^{2} to get -6x^{2}.

$$-6x^{2}+12x+9=-162x+729$$

Add 162x to both sides.

$$-6x^{2}+12x+9+162x=729$$

Combine 12x and 162x to get 174x.

$$-6x^{2}+174x+9=729$$

Subtract 729 from both sides.

$$-6x^{2}+174x+9-729=0$$

Subtract 729 from 9 to get -720.

$$-6x^{2}+174x-720=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -6 for a, 174 for b, and -720 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-174±\sqrt{174^{2}-4\left(-6\right)\left(-720\right)}}{2\left(-6\right)}$$

Square 174.

$$x=\frac{-174±\sqrt{30276-4\left(-6\right)\left(-720\right)}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$x=\frac{-174±\sqrt{30276+24\left(-720\right)}}{2\left(-6\right)}$$

Multiply 24 times -720.

$$x=\frac{-174±\sqrt{30276-17280}}{2\left(-6\right)}$$

Add 30276 to -17280.

$$x=\frac{-174±\sqrt{12996}}{2\left(-6\right)}$$

Take the square root of 12996.

$$x=\frac{-174±114}{2\left(-6\right)}$$

Multiply 2 times -6.

$$x=\frac{-174±114}{-12}$$

Now solve the equation x=\frac{-174±114}{-12} when ± is plus. Add -174 to 114.

$$x=-\frac{60}{-12}$$

Divide -60 by -12.

$$x=5$$

Now solve the equation x=\frac{-174±114}{-12} when ± is minus. Subtract 114 from -174.

$$x=-\frac{288}{-12}$$

Divide -288 by -12.

$$x=24$$

The equation is now solved.

$$x=5 x=24$$

Steps for Completing the Square

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

$$4x^{2}+12x+9=x^{2}+\left(x+27-4x\right)^{2}$$

Combine x and -4x to get -3x.

$$4x^{2}+12x+9=x^{2}+\left(-3x+27\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-3x+27\right)^{2}.

$$4x^{2}+12x+9=x^{2}+9x^{2}-162x+729$$

Combine x^{2} and 9x^{2} to get 10x^{2}.

$$4x^{2}+12x+9=10x^{2}-162x+729$$

Subtract 10x^{2} from both sides.

$$4x^{2}+12x+9-10x^{2}=-162x+729$$

Combine 4x^{2} and -10x^{2} to get -6x^{2}.

$$-6x^{2}+12x+9=-162x+729$$

Add 162x to both sides.

$$-6x^{2}+12x+9+162x=729$$

Combine 12x and 162x to get 174x.

$$-6x^{2}+174x+9=729$$

Subtract 9 from both sides.

$$-6x^{2}+174x=729-9$$

Subtract 9 from 729 to get 720.

$$-6x^{2}+174x=720$$

Divide both sides by -6.

$$\frac{-6x^{2}+174x}{-6}=\frac{720}{-6}$$

Dividing by -6 undoes the multiplication by -6.

$$x^{2}+\frac{174}{-6}x=\frac{720}{-6}$$

Divide 174 by -6.

$$x^{2}-29x=\frac{720}{-6}$$

Divide 720 by -6.

$$x^{2}-29x=-120$$

Divide -29, the coefficient of the x term, by 2 to get -\frac{29}{2}. Then add the square of -\frac{29}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-29x+\left(-\frac{29}{2}\right)^{2}=-120+\left(-\frac{29}{2}\right)^{2}$$

Square -\frac{29}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-29x+\frac{841}{4}=-120+\frac{841}{4}$$

Add -120 to \frac{841}{4}.

$$x^{2}-29x+\frac{841}{4}=\frac{361}{4}$$

Factor x^{2}-29x+\frac{841}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{29}{2}\right)^{2}=\frac{361}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{29}{2}\right)^{2}}=\sqrt{\frac{361}{4}}$$

Simplify.

$$x-\frac{29}{2}=\frac{19}{2} x-\frac{29}{2}=-\frac{19}{2}$$

Add \frac{29}{2} to both sides of the equation.

$$x=24 x=5$$