Solve 6^2+(2*5+x)^2=4^2-(2*5-x)^2

$$6 ^ { 2 } + ( 2 \cdot 5 + x ) ^ { 2 } = 4 ^ { 2 } - ( 2 \cdot 5 - x ) ^ { 2 }$$

Solve for x (complex solution)

$$x=-\sqrt{110}i\approx -0-10.488088482i$$

$$x=\sqrt{110}i\approx 10.488088482i$$

Steps by Finding Square Root

Calculate 6 to the power of 2 and get 36.

$$36+\left(2\times 5+x\right)^{2}=4^{2}-\left(2\times 5-x\right)^{2}$$

Multiply 2 and 5 to get 10.

$$36+\left(10+x\right)^{2}=4^{2}-\left(2\times 5-x\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(10+x\right)^{2}.

$$36+100+20x+x^{2}=4^{2}-\left(2\times 5-x\right)^{2}$$

Add 36 and 100 to get 136.

$$136+20x+x^{2}=4^{2}-\left(2\times 5-x\right)^{2}$$

Calculate 4 to the power of 2 and get 16.

$$136+20x+x^{2}=16-\left(2\times 5-x\right)^{2}$$

Multiply 2 and 5 to get 10.

$$136+20x+x^{2}=16-\left(10-x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(10-x\right)^{2}.

$$136+20x+x^{2}=16-\left(100-20x+x^{2}\right)$$

To find the opposite of 100-20x+x^{2}, find the opposite of each term.

$$136+20x+x^{2}=16-100+20x-x^{2}$$

Subtract 100 from 16 to get -84.

$$136+20x+x^{2}=-84+20x-x^{2}$$

Subtract 20x from both sides.

$$136+20x+x^{2}-20x=-84-x^{2}$$

Combine 20x and -20x to get 0.

$$136+x^{2}=-84-x^{2}$$

Add x^{2} to both sides.

$$136+x^{2}+x^{2}=-84$$

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

$$136+2x^{2}=-84$$

Subtract 136 from both sides.

$$2x^{2}=-84-136$$

Subtract 136 from -84 to get -220.

$$2x^{2}=-220$$

Divide both sides by 2.

$$x^{2}=\frac{-220}{2}$$

Divide -220 by 2 to get -110.

$$x^{2}=-110$$

The equation is now solved.

$$x=\sqrt{110}i x=-\sqrt{110}i$$

Steps Using the Quadratic Formula

Calculate 6 to the power of 2 and get 36.

$$36+\left(2\times 5+x\right)^{2}=4^{2}-\left(2\times 5-x\right)^{2}$$

Multiply 2 and 5 to get 10.

$$36+\left(10+x\right)^{2}=4^{2}-\left(2\times 5-x\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(10+x\right)^{2}.

$$36+100+20x+x^{2}=4^{2}-\left(2\times 5-x\right)^{2}$$

Add 36 and 100 to get 136.

$$136+20x+x^{2}=4^{2}-\left(2\times 5-x\right)^{2}$$

Calculate 4 to the power of 2 and get 16.

$$136+20x+x^{2}=16-\left(2\times 5-x\right)^{2}$$

Multiply 2 and 5 to get 10.

$$136+20x+x^{2}=16-\left(10-x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(10-x\right)^{2}.

$$136+20x+x^{2}=16-\left(100-20x+x^{2}\right)$$

To find the opposite of 100-20x+x^{2}, find the opposite of each term.

$$136+20x+x^{2}=16-100+20x-x^{2}$$

Subtract 100 from 16 to get -84.

$$136+20x+x^{2}=-84+20x-x^{2}$$

Subtract -84 from both sides.

$$136+20x+x^{2}-\left(-84\right)=20x-x^{2}$$

The opposite of -84 is 84.

$$136+20x+x^{2}+84=20x-x^{2}$$

Subtract 20x from both sides.

$$136+20x+x^{2}+84-20x=-x^{2}$$

Add 136 and 84 to get 220.

$$220+20x+x^{2}-20x=-x^{2}$$

Combine 20x and -20x to get 0.

$$220+x^{2}=-x^{2}$$

Add x^{2} to both sides.

$$220+x^{2}+x^{2}=0$$

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

$$220+2x^{2}=0$$

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.

$$2x^{2}+220=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 0 for b, and 220 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 2\times 220}}{2\times 2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 2\times 220}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{0±\sqrt{-8\times 220}}{2\times 2}$$

Multiply -8 times 220.

$$x=\frac{0±\sqrt{-1760}}{2\times 2}$$

Take the square root of -1760.

$$x=\frac{0±4\sqrt{110}i}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{0±4\sqrt{110}i}{4}$$

Now solve the equation x=\frac{0±4\sqrt{110}i}{4} when ± is plus.

$$x=\sqrt{110}i$$

Now solve the equation x=\frac{0±4\sqrt{110}i}{4} when ± is minus.

$$x=-\sqrt{110}i$$

The equation is now solved.

$$x=\sqrt{110}i x=-\sqrt{110}i$$