Calculate $\sqrt{x}$ to the power of $2$ and get $x$.
$$1x=\left(10-5x\right)^{2}$$
Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(10-5x\right)^{2}$.
$$1x=100-100x+25x^{2}$$
Add $100x$ to both sides.
$$1x+100x=100+25x^{2}$$
Combine $1x$ and $100x$ to get $101x$.
$$101x=100+25x^{2}$$
Subtract $25x^{2}$ from both sides.
$$101x-25x^{2}=100$$
All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.
$$-25x^{2}+101x=100$$
Subtract $100$ from both sides of the equation.
$$-25x^{2}+101x-100=100-100$$
Subtracting $100$ from itself leaves $0$.
$$-25x^{2}+101x-100=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-25$ for $a$, $101$ for $b$, and $-100$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
