Krdytkyu

Make a the subject,
$$sqrt{c-a^2}=a+btag1$$

$$c-a^2=(a+b)^2tag2$$

$$c-a^2=a^2+b^2tag3$$

$$c-b^2=a^2+a^2tag4$$

$$a^2=frac{c-b^2}{2}tag5$$

$$a=sqrt{frac{c-b^2}{2}}tag6$$

My teacher said I am wrong!

I trace all my steps but can't see where I went wrong.

You expanded incorrectly in $(3)$. Recall

$$color{blue}{(a+b)^2 = a^2+2ab+b^2} color{red}{neq a^2+b^2}$$

You can try expanding it to confirm.

$$(a+b)^2 = (a+b)(a+b) = a^2+ab+ba+b^2 = a^2+2ab+b^2$$

So you can continue from $(2)$.

$$c-a^2 = (a+b)^2$$

$$c^2-a^2 = a^2+2ab+b^2$$

$$0 = 2a^2+2ab+b^2-c^2$$

You want to solve for $a$, so group the terms for solving more easily.

$$0 = 2a^2+2ba+big(b^2-c^2big)$$

Use the Quadratic Formula.

$$color{blue}{a}x^2+color{purple}{b}x+c = 0 implies x = frac{-color{purple}{b}pmsqrt{color{purple}{b}^2-4color{blue}{a}c}}{2color{blue}{a}}$$

$$0 = color{blue}{2}a^2+color{purple}{2b}a+big(b^2-c^2big) implies a = frac{-color{purple}{2b}pmsqrt{(color{purple}{2b})^2-4(color{blue}{2})big(b^2-c^2big)}}{2(color{blue}{2})}$$

Simplifying, you get

$$a = frac{-2bpmsqrt{4b^2-8b^2+8c^2}}{4}$$

$$a = frac{-2bpmsqrt{8c^2-4b^2}}{4}$$

$$a = frac{-2bpmsqrt{4(2c^2-b^2)}}{4}$$

$$a = frac{-2bpm 2sqrt{2c^2-b^2}}{4}$$

You can also complete the square, which yields the same answer.

Note that $$(a+b)^2 = a^2+b^2 + 2ab$$

After you obtain a quadratic equation in $a$, you might like to use the quadratic formula and verify if every solution that you obtain are indeed a solution.

(3) is defective:
$49=(3+4)^2neq3^2+4^2$. We can’t pretend that Pythagoras never lived.