Evaluate: (6m+1)^2-5(6m+1)+4=0

Expression: ${\left( 6m+1 \right)}^{2}-5\left( 6m+1 \right)+4=0$

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

$36{m}^{2}+12m+1-5\left( 6m+1 \right)+4=0$

Distribute $-5$ through the parentheses

$36{m}^{2}+12m+1-30m-5+4=0$

Collect like terms

$36{m}^{2}-18m+1-5+4=0$

Calculate the sum or difference

$36{m}^{2}-18m+0=0$

Removing $0$ doesn't change the value, so remove it from the expression

$36{m}^{2}-18m=0$

Write the quadratic equation in the appropriate form

${m}^{2}-\frac{ 1 }{ 2 }m=0$

Identify the coefficients $p$ and $q$ of the quadratic equation

$\begin{array} { l }p=-\frac{ 1 }{ 2 },& q=0\end{array}$

Substitute $p=-\frac{ 1 }{ 2 }$ and $q=0$ into the PQ formula $x=-\frac{ p }{ 2 }\pm\sqrt{ {\left( \frac{ p }{ 2 } \right)}^{2}-q }$

$m=-\frac{ -\frac{ 1 }{ 2 } }{ 2 }\pm\sqrt{ {\left( \frac{ -\frac{ 1 }{ 2 } }{ 2 } \right)}^{2}-0 }$

Removing $0$ doesn't change the value, so remove it from the expression

$m=-\frac{ -\frac{ 1 }{ 2 } }{ 2 }\pm\sqrt{ {\left( \frac{ -\frac{ 1 }{ 2 } }{ 2 } \right)}^{2} }$

Determine the sign of the fraction

$m=\frac{ \frac{ 1 }{ 2 } }{ 2 }\pm\sqrt{ {\left( \frac{ -\frac{ 1 }{ 2 } }{ 2 } \right)}^{2} }$

Use $\frac{ -a }{ b }=\frac{ a }{ -b }=-\frac{ a }{ b }$ to rewrite the fraction

$m=\frac{ \frac{ 1 }{ 2 } }{ 2 }\pm\sqrt{ {\left( -\frac{ \frac{ 1 }{ 2 } }{ 2 } \right)}^{2} }$

Simplify the complex fraction

$m=\frac{ 1 }{ 4 }\pm\sqrt{ {\left( -\frac{ \frac{ 1 }{ 2 } }{ 2 } \right)}^{2} }$

Simplify the complex fraction

$m=\frac{ 1 }{ 4 }\pm\sqrt{ {\left( -\frac{ 1 }{ 4 } \right)}^{2} }$

Reduce the index of the radical and exponent with $2$

$m=\frac{ 1 }{ 4 }\pm\frac{ 1 }{ 4 }$

Write the solutions, one with a $+$ sign and one with a $-$ sign

$\begin{array} { l }m=\frac{ 1 }{ 4 }+\frac{ 1 }{ 4 },\\m=\frac{ 1 }{ 4 }-\frac{ 1 }{ 4 }\end{array}$

Add the fractions

$\begin{array} { l }m=\frac{ 1 }{ 2 },\\m=\frac{ 1 }{ 4 }-\frac{ 1 }{ 4 }\end{array}$

Subtract the fractions

$\begin{array} { l }m=\frac{ 1 }{ 2 },\\m=0\end{array}$

The equation has $2$ solutions

$\begin{align*}&\begin{array} { l }m_1=0,& m_2=\frac{ 1 }{ 2 }\end{array} \\&\begin{array} { l }m_1=0,& m_2=0.5\end{array}\end{align*}$