Solving a quadratic equation using the quadratic formulaSample question

\(x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\)

\(x = \frac{{ - 7 \pm \sqrt {{7^2} - (4 \times 3 \times 4)} }}{{2 \times 3}}\)

\(x = \frac{{ - 7 \pm \sqrt {49 - (48)} }}{6}\)

\(x = \frac{{ - 7 \pm \sqrt 1 }}{6}\)

\(x = \frac{{ - 7 + \sqrt 1 }}{6}\)

\(x=\frac{-6}{6}\)

\(x=-1\)

\(x = \frac{{ - 7 - \sqrt 1 }}{6}\)

\(x = - 1.33\,(to\,2\,d.p.)\)

Therefore \(x = - 1\,and\,x = - 1.33\)