Evaluate: (10c-9)^2=12^2

Expression: ${\left( 10c-9 \right)}^{2}={12}^{2}$

Move the constant to the left-hand side and change its sign

${\left( 10c-9 \right)}^{2}-{12}^{2}=0$

Use ${\left( a-b \right)}^{2}={a}^{2}-2ab+{b}^{2}$ to expand the expression

$100{c}^{2}-180c+81-{12}^{2}=0$

Evaluate the power

$100{c}^{2}-180c+81-144=0$

Write $-180c$ as a difference

$100{c}^{2}+30c-210c+81-144=0$

Calculate the difference

$100{c}^{2}+30c-210c-63=0$

Factor out $10c$ from the expression

$10c \times \left( 10c+3 \right)-210c-63=0$

Factor out $-21$ from the expression

$10c \times \left( 10c+3 \right)-21\left( 10c+3 \right)=0$

Factor out $10c+3$ from the expression

$\left( 10c+3 \right) \times \left( 10c-21 \right)=0$

When the product of factors equals $0$, at least one factor is $0$

$\begin{array} { l }10c+3=0,\\10c-21=0\end{array}$

Solve the equation for $c$

$\begin{array} { l }c=-\frac{ 3 }{ 10 },\\10c-21=0\end{array}$

Solve the equation for $c$

$\begin{array} { l }c=-\frac{ 3 }{ 10 },\\c=\frac{ 21 }{ 10 }\end{array}$

The equation has $2$ solutions

$\begin{align*}&\begin{array} { l }c_1=-\frac{ 3 }{ 10 },& c_2=\frac{ 21 }{ 10 }\end{array} \\&\begin{array} { l }c_1=-0.3,& c_2=2.1\end{array}\end{align*}$