Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(x-2\right)^{2}$.
$$x^{2}-4x+4=100$$
Subtract $100$ from both sides.
$$x^{2}-4x+4-100=0$$
Subtract $100$ from $4$ to get $-96$.
$$x^{2}-4x-96=0$$
To solve the equation, factor $x^{2}-4x-96$ using formula $x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right)$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-4$$ $$ab=-96$$
Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product $-96$.
Rewrite factored expression $\left(x+a\right)\left(x+b\right)$ using the obtained values.
$$\left(x-12\right)\left(x+8\right)$$
To find equation solutions, solve $x-12=0$ and $x+8=0$.
$$x=12$$ $$x=-8$$
Steps Using Factoring By Grouping
Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(x-2\right)^{2}$.
$$x^{2}-4x+4=100$$
Subtract $100$ from both sides.
$$x^{2}-4x+4-100=0$$
Subtract $100$ from $4$ to get $-96$.
$$x^{2}-4x-96=0$$
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $x^{2}+ax+bx-96$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-4$$ $$ab=1\left(-96\right)=-96$$
Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product $-96$.
Rewrite $x^{2}-4x-96$ as $\left(x^{2}-12x\right)+\left(8x-96\right)$.
$$\left(x^{2}-12x\right)+\left(8x-96\right)$$
Factor out $x$ in the first and $8$ in the second group.
$$x\left(x-12\right)+8\left(x-12\right)$$
Factor out common term $x-12$ by using distributive property.
$$\left(x-12\right)\left(x+8\right)$$
To find equation solutions, solve $x-12=0$ and $x+8=0$.
$$x=12$$ $$x=-8$$
Steps Using the Quadratic Formula
Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(x-2\right)^{2}$.
$$x^{2}-4x+4=100$$
Subtract $100$ from both sides.
$$x^{2}-4x+4-100=0$$
Subtract $100$ from $4$ to get $-96$.
$$x^{2}-4x-96=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $-4$ for $b$, and $-96$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
