Example  1
Solve the equation \(2x^2 + 6x + 1 = 0\) using the quadratic formula, and express the roots as exact values.

Step 1: Substitute the values of \(a\), \(b\), and \(c\) into the quadratic formula.

\(\begin{align}
 a &= 2  \\
 b &= 6  \\
 c &= 1  \end{align}\)

\[\begin{array}
 x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}} \\
 x = \frac{{ - 6 \pm \sqrt {6^2 - 4\left( 2 \right)\left( 1 \right)} }}{{2\left( 2 \right)}} \\
 \end{array}\]

Step 2: Simplify.

\[\begin{align} 
x &= \frac{{ -6 \pm \sqrt {36 - 8} }}{4} \\
 x &= \frac{{ -6 \pm \sqrt {28} }}{4} \\
 x &= \frac{{ -6 \pm 2\sqrt 7 }}{4} \\
 x &= \frac{{ -3 \pm \sqrt 7 }}{2}  \end{align}\]



Because all three terms have a GCF of \(2\), simplify by dividing all three terms by \(2\).

Step 3: State each root individually, as an exact value.

\(x = \frac{{ - 3 + \sqrt 7 }}{2}\) and \(x = \frac{{ - 3 - \sqrt 7 }}{2}\)

Step 4: Verify by substituting the solutions into the original equation.

For \(x = \frac{{ - 3 + \sqrt 7 }}{2}\),

Left Side
 Right Side
\(\begin{array}{r}
 2x^2 + 6x + 1 \\
 2\left( {\frac{{ - 3 + \sqrt 7 }}{2}} \right)^2 + 6\left( {\frac{{ - 3 + \sqrt 7 }}{2}} \right) + 1 \\
 2\left[ {\frac{{\left( { - 3 + \sqrt 7 } \right)^2 }}{4}} \right] + 3\left( { - 3 + \sqrt 7 } \right) + 1 \\
 2\left[ {\frac{{\left( {9 - 3\sqrt 7 - 3\sqrt 7 + 7} \right)}}{4}} \right] + 3\left( { - 3 + \sqrt 7 } \right) + 1 \\
 \frac{{\left( {9 - 6\sqrt 7  + 7} \right)}}{2} - 9 + 3\sqrt 7 + 1 \\
 \frac{{16 - 6\sqrt 7 }}{2} - 8 + 3\sqrt 7  \\
 8 - 3\sqrt 7 - 8 + 3\sqrt 7  \\
 0  \end{array}\)
 \(0\)
LS = RS             

For \(x = \frac{{ - 3 - \sqrt 7 }}{2}\)

Left Side
 Right Side
\(\begin{array}{r}
 2x^2 + 6x + 1 \\
 2\left( {\frac{{ - 3 - \sqrt 7 }}{2}} \right)^2 + 6\left( {\frac{{ - 3 - \sqrt 7 }}{2}} \right) + 1 \\
 2\left[ {\frac{{\left( { - 3 - \sqrt 7 } \right)^2 }}{4}} \right] + 3\left( { - 3 - \sqrt 7 } \right) + 1 \\
 2\left[ {\frac{{\left( {9 + 3\sqrt 7  + 3\sqrt 7 + 7} \right)}}{4}} \right] + 3\left( { - 3 - \sqrt 7 } \right) + 1 \\
 \frac{{\left( {9 + 6\sqrt 7 + 7} \right)}}{2} - 9 - 3\sqrt 7  + 1 \\
 \frac{{16 + 6\sqrt 7 }}{2} - 8 - 3\sqrt 7  \\
 8 + 3\sqrt 7 - 8 - 3\sqrt 7  \\
 0  \end{array}\)
 \(0\)
LS = RS             

The solutions to the quadratic equation \(2x^2 + 6x + 1 = 0\) are \(x = \frac{{ - 3 + \sqrt 7 }}{2}\) and \(x = \frac{{ - 3 - \sqrt 7 }}{2}\).