Solve x^2+16x+64=75-3x^2

$${ x }^{ 2 } +16x+64 = 75-3 { x }^{ 2 }$$

Solve for x

$$x=\frac{3\sqrt{3}}{2}-2\approx 0.598076211$$

$$x=-\frac{3\sqrt{3}}{2}-2\approx -4.598076211$$

Steps Using the Quadratic Formula

Subtract 75 from both sides.

$$x^{2}+16x+64-75=-3x^{2}$$

Subtract 75 from 64 to get -11.

$$x^{2}+16x-11=-3x^{2}$$

Add 3x^{2} to both sides.

$$x^{2}+16x-11+3x^{2}=0$$

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

$$4x^{2}+16x-11=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 16 for b, and -11 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-16±\sqrt{16^{2}-4\times 4\left(-11\right)}}{2\times 4}$$

Square 16.

$$x=\frac{-16±\sqrt{256-4\times 4\left(-11\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-16±\sqrt{256-16\left(-11\right)}}{2\times 4}$$

Multiply -16 times -11.

$$x=\frac{-16±\sqrt{256+176}}{2\times 4}$$

Add 256 to 176.

$$x=\frac{-16±\sqrt{432}}{2\times 4}$$

Take the square root of 432.

$$x=\frac{-16±12\sqrt{3}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-16±12\sqrt{3}}{8}$$

Now solve the equation x=\frac{-16±12\sqrt{3}}{8} when ± is plus. Add -16 to 12\sqrt{3}.

$$x=\frac{12\sqrt{3}-16}{8}$$

Divide -16+12\sqrt{3} by 8.

$$x=\frac{3\sqrt{3}}{2}-2$$

Now solve the equation x=\frac{-16±12\sqrt{3}}{8} when ± is minus. Subtract 12\sqrt{3} from -16.

$$x=\frac{-12\sqrt{3}-16}{8}$$

Divide -16-12\sqrt{3} by 8.

$$x=-\frac{3\sqrt{3}}{2}-2$$

The equation is now solved.

$$x=\frac{3\sqrt{3}}{2}-2 x=-\frac{3\sqrt{3}}{2}-2$$

Steps for Completing the Square

Add 3x^{2} to both sides.

$$x^{2}+16x+64+3x^{2}=75$$

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

$$4x^{2}+16x+64=75$$

Subtract 64 from both sides.

$$4x^{2}+16x=75-64$$

Subtract 64 from 75 to get 11.

$$4x^{2}+16x=11$$

Divide both sides by 4.

$$\frac{4x^{2}+16x}{4}=\frac{11}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{16}{4}x=\frac{11}{4}$$

Divide 16 by 4.

$$x^{2}+4x=\frac{11}{4}$$

Divide 4, the coefficient of the x term, by 2 to get 2. Then add the square of 2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+4x+2^{2}=\frac{11}{4}+2^{2}$$

Square 2.

$$x^{2}+4x+4=\frac{11}{4}+4$$

Add \frac{11}{4} to 4.

$$x^{2}+4x+4=\frac{27}{4}$$

Factor x^{2}+4x+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}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{\frac{27}{4}}$$

Simplify.

$$x+2=\frac{3\sqrt{3}}{2} x+2=-\frac{3\sqrt{3}}{2}$$

Subtract 2 from both sides of the equation.

$$x=\frac{3\sqrt{3}}{2}-2 x=-\frac{3\sqrt{3}}{2}-2$$