Solve 5^2-(5-x^2)=6^2-x^2

$${ 5 }^{ 2 } -(5- { x }^{ 2 } )= { 6 }^{ 2 } - { x }^{ 2 }$$

Solve for x

$$x=2\sqrt{2}\approx 2.828427125$$

$$x=-2\sqrt{2}\approx -2.828427125$$

Steps by Finding Square Root

Calculate 5 to the power of 2 and get 25.

$$25-\left(5-x^{2}\right)=6^{2}-x^{2}$$

To find the opposite of 5-x^{2}, find the opposite of each term.

$$25-5+x^{2}=6^{2}-x^{2}$$

Subtract 5 from 25 to get 20.

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

Calculate 6 to the power of 2 and get 36.

$$20+x^{2}=36-x^{2}$$

Add x^{2} to both sides.

$$20+x^{2}+x^{2}=36$$

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

$$20+2x^{2}=36$$

Subtract 20 from both sides.

$$2x^{2}=36-20$$

Subtract 20 from 36 to get 16.

$$2x^{2}=16$$

Divide both sides by 2.

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

Divide 16 by 2 to get 8.

$$x^{2}=8$$

Take the square root of both sides of the equation.

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

Steps Using the Quadratic Formula

Calculate 5 to the power of 2 and get 25.

$$25-\left(5-x^{2}\right)=6^{2}-x^{2}$$

To find the opposite of 5-x^{2}, find the opposite of each term.

$$25-5+x^{2}=6^{2}-x^{2}$$

Subtract 5 from 25 to get 20.

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

Calculate 6 to the power of 2 and get 36.

$$20+x^{2}=36-x^{2}$$

Subtract 36 from both sides.

$$20+x^{2}-36=-x^{2}$$

Subtract 36 from 20 to get -16.

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

Add x^{2} to both sides.

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

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

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

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.

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

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

$$x=\frac{0±\sqrt{0^{2}-4\times 2\left(-16\right)}}{2\times 2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 2\left(-16\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{0±\sqrt{-8\left(-16\right)}}{2\times 2}$$

Multiply -8 times -16.

$$x=\frac{0±\sqrt{128}}{2\times 2}$$

Take the square root of 128.

$$x=\frac{0±8\sqrt{2}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{0±8\sqrt{2}}{4}$$

Now solve the equation x=\frac{0±8\sqrt{2}}{4} when ± is plus.

$$x=2\sqrt{2}$$

Now solve the equation x=\frac{0±8\sqrt{2}}{4} when ± is minus.

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

The equation is now solved.

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