Solve 4+x^2-64=100-x^2

$$4+ { x }^{ 2 } -64=100- { x }^{ 2 }$$

Solve for x

$$x=4\sqrt{5}\approx 8.94427191$$

$$x=-4\sqrt{5}\approx -8.94427191$$

Steps by Finding Square Root

Subtract 64 from 4 to get -60.

$$-60+x^{2}=100-x^{2}$$

Add x^{2} to both sides.

$$-60+x^{2}+x^{2}=100$$

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

$$-60+2x^{2}=100$$

Add 60 to both sides.

$$2x^{2}=100+60$$

Add 100 and 60 to get 160.

$$2x^{2}=160$$

Divide both sides by 2.

$$x^{2}=\frac{160}{2}$$

Divide 160 by 2 to get 80.

$$x^{2}=80$$

Take the square root of both sides of the equation.

$$x=4\sqrt{5} x=-4\sqrt{5}$$

Steps Using the Quadratic Formula

Subtract 64 from 4 to get -60.

$$-60+x^{2}=100-x^{2}$$

Subtract 100 from both sides.

$$-60+x^{2}-100=-x^{2}$$

Subtract 100 from -60 to get -160.

$$-160+x^{2}=-x^{2}$$

Add x^{2} to both sides.

$$-160+x^{2}+x^{2}=0$$

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

$$-160+2x^{2}=0$$

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

$$2x^{2}-160=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 0 for b, and -160 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 2\left(-160\right)}}{2\times 2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 2\left(-160\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{0±\sqrt{-8\left(-160\right)}}{2\times 2}$$

Multiply -8 times -160.

$$x=\frac{0±\sqrt{1280}}{2\times 2}$$

Take the square root of 1280.

$$x=\frac{0±16\sqrt{5}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{0±16\sqrt{5}}{4}$$

Now solve the equation x=\frac{0±16\sqrt{5}}{4} when ± is plus.

$$x=4\sqrt{5}$$

Now solve the equation x=\frac{0±16\sqrt{5}}{4} when ± is minus.

$$x=-4\sqrt{5}$$

The equation is now solved.

$$x=4\sqrt{5} x=-4\sqrt{5}$$