Solving a quadratic equation using the quadratic formulaWorked example

Solve \(2{x^2} - 5x - 6 = 0\)

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

\(x = \frac{{ - ( - 5) \pm \sqrt {{{( - 5)}^2} - (4 \times 2 \times ( - 6))} }}{{2 \times 2}}\)

\(x = \frac{{5 \pm \sqrt {25 - ( - 48)} }}{4}\)

\(x = \frac{{5 \pm \sqrt {25 + 48} }}{4}\)

\(x = \frac{{5 \pm \sqrt {73} }}{4}\)

\(x = \frac{{5 + \sqrt {73} }}{4}\)

\(x=\frac{13.544}{4}\)

\(x = 3.39\,(to\,2\,d.p.)\)

\(x = \frac{{5 - \sqrt {73} }}{4}\)

\(x=\frac{-3.544}{4}\)

\(x = - 0.89\,(to\,2\,d.p.)\)

Therefore \(x = 3.39\,and\,x = - 0.89\)