Solve 4x^2+9=left(9-xright)^2

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

Solve for x (complex solution)

$$x=\sqrt{33}-3\approx 2.744562647$$

$$x=-\left(\sqrt{33}+3\right)\approx -8.744562647$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(9-x\right)^{2}.

$$4x^{2}+9=81-18x+x^{2}$$

Subtract 81 from both sides.

$$4x^{2}+9-81=-18x+x^{2}$$

Subtract 81 from 9 to get -72.

$$4x^{2}-72=-18x+x^{2}$$

Add 18x to both sides.

$$4x^{2}-72+18x=x^{2}$$

Subtract x^{2} from both sides.

$$4x^{2}-72+18x-x^{2}=0$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}-72+18x=0$$

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.

$$3x^{2}+18x-72=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 18 for b, and -72 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-18±\sqrt{18^{2}-4\times 3\left(-72\right)}}{2\times 3}$$

Square 18.

$$x=\frac{-18±\sqrt{324-4\times 3\left(-72\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-18±\sqrt{324-12\left(-72\right)}}{2\times 3}$$

Multiply -12 times -72.

$$x=\frac{-18±\sqrt{324+864}}{2\times 3}$$

Add 324 to 864.

$$x=\frac{-18±\sqrt{1188}}{2\times 3}$$

Take the square root of 1188.

$$x=\frac{-18±6\sqrt{33}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-18±6\sqrt{33}}{6}$$

Now solve the equation x=\frac{-18±6\sqrt{33}}{6} when ± is plus. Add -18 to 6\sqrt{33}.

$$x=\frac{6\sqrt{33}-18}{6}$$

Divide -18+6\sqrt{33} by 6.

$$x=\sqrt{33}-3$$

Now solve the equation x=\frac{-18±6\sqrt{33}}{6} when ± is minus. Subtract 6\sqrt{33} from -18.

$$x=\frac{-6\sqrt{33}-18}{6}$$

Divide -18-6\sqrt{33} by 6.

$$x=-\sqrt{33}-3$$

The equation is now solved.

$$x=\sqrt{33}-3 x=-\sqrt{33}-3$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(9-x\right)^{2}.

$$4x^{2}+9=81-18x+x^{2}$$

Add 18x to both sides.

$$4x^{2}+9+18x=81+x^{2}$$

Subtract x^{2} from both sides.

$$4x^{2}+9+18x-x^{2}=81$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}+9+18x=81$$

Subtract 9 from both sides.

$$3x^{2}+18x=81-9$$

Subtract 9 from 81 to get 72.

$$3x^{2}+18x=72$$

Divide both sides by 3.

$$\frac{3x^{2}+18x}{3}=\frac{72}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{18}{3}x=\frac{72}{3}$$

Divide 18 by 3.

$$x^{2}+6x=\frac{72}{3}$$

Divide 72 by 3.

$$x^{2}+6x=24$$

Divide 6, the coefficient of the x term, by 2 to get 3. Then add the square of 3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+6x+3^{2}=24+3^{2}$$

Square 3.

$$x^{2}+6x+9=24+9$$

Add 24 to 9.

$$x^{2}+6x+9=33$$

Factor x^{2}+6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+3\right)^{2}=33$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+3\right)^{2}}=\sqrt{33}$$

Simplify.

$$x+3=\sqrt{33} x+3=-\sqrt{33}$$

Subtract 3 from both sides of the equation.

$$x=\sqrt{33}-3 x=-\sqrt{33}-3$$

Solve for x

$$x=\sqrt{33}-3\approx 2.744562647$$

$$x=-\sqrt{33}-3\approx -8.744562647$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(9-x\right)^{2}.

$$4x^{2}+9=81-18x+x^{2}$$

Subtract 81 from both sides.

$$4x^{2}+9-81=-18x+x^{2}$$

Subtract 81 from 9 to get -72.

$$4x^{2}-72=-18x+x^{2}$$

Add 18x to both sides.

$$4x^{2}-72+18x=x^{2}$$

Subtract x^{2} from both sides.

$$4x^{2}-72+18x-x^{2}=0$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}-72+18x=0$$

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.

$$3x^{2}+18x-72=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 18 for b, and -72 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-18±\sqrt{18^{2}-4\times 3\left(-72\right)}}{2\times 3}$$

Square 18.

$$x=\frac{-18±\sqrt{324-4\times 3\left(-72\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-18±\sqrt{324-12\left(-72\right)}}{2\times 3}$$

Multiply -12 times -72.

$$x=\frac{-18±\sqrt{324+864}}{2\times 3}$$

Add 324 to 864.

$$x=\frac{-18±\sqrt{1188}}{2\times 3}$$

Take the square root of 1188.

$$x=\frac{-18±6\sqrt{33}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-18±6\sqrt{33}}{6}$$

Now solve the equation x=\frac{-18±6\sqrt{33}}{6} when ± is plus. Add -18 to 6\sqrt{33}.

$$x=\frac{6\sqrt{33}-18}{6}$$

Divide -18+6\sqrt{33} by 6.

$$x=\sqrt{33}-3$$

Now solve the equation x=\frac{-18±6\sqrt{33}}{6} when ± is minus. Subtract 6\sqrt{33} from -18.

$$x=\frac{-6\sqrt{33}-18}{6}$$

Divide -18-6\sqrt{33} by 6.

$$x=-\sqrt{33}-3$$

The equation is now solved.

$$x=\sqrt{33}-3 x=-\sqrt{33}-3$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(9-x\right)^{2}.

$$4x^{2}+9=81-18x+x^{2}$$

Add 18x to both sides.

$$4x^{2}+9+18x=81+x^{2}$$

Subtract x^{2} from both sides.

$$4x^{2}+9+18x-x^{2}=81$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}+9+18x=81$$

Subtract 9 from both sides.

$$3x^{2}+18x=81-9$$

Subtract 9 from 81 to get 72.

$$3x^{2}+18x=72$$

Divide both sides by 3.

$$\frac{3x^{2}+18x}{3}=\frac{72}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{18}{3}x=\frac{72}{3}$$

Divide 18 by 3.

$$x^{2}+6x=\frac{72}{3}$$

Divide 72 by 3.

$$x^{2}+6x=24$$

Divide 6, the coefficient of the x term, by 2 to get 3. Then add the square of 3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+6x+3^{2}=24+3^{2}$$

Square 3.

$$x^{2}+6x+9=24+9$$

Add 24 to 9.

$$x^{2}+6x+9=33$$

Factor x^{2}+6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+3\right)^{2}=33$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+3\right)^{2}}=\sqrt{33}$$

Simplify.

$$x+3=\sqrt{33} x+3=-\sqrt{33}$$

Subtract 3 from both sides of the equation.

$$x=\sqrt{33}-3 x=-\sqrt{33}-3$$