# How do you complete the square to solve 4x^2 - 7x - 2 = 0?

May 23, 2015

In this way:

$4 {x}^{2} - 7 x - 2 = 0 \Rightarrow 4 \left({x}^{2} - \frac{7}{4} x\right) - 2 = 0 \Rightarrow$

$4 \left({x}^{2} - \frac{7}{4} x + \frac{49}{64} - \frac{49}{64}\right) - 2 = 0 \Rightarrow$

$4 \left({x}^{2} - \frac{7}{4} x + \frac{49}{64}\right) - \frac{49}{16} - 2 = 0 \Rightarrow$

$4 {\left(x - \frac{7}{8}\right)}^{2} = \frac{49}{16} + 2 \Rightarrow 4 {\left(x - \frac{7}{8}\right)}^{2} = \frac{49 + 32}{16} \Rightarrow$

${\left(x - \frac{7}{8}\right)}^{2} = \frac{81}{16} \cdot \frac{1}{4} \Rightarrow {\left(x - \frac{7}{8}\right)}^{2} = \frac{81}{64} \Rightarrow$

$x - \frac{7}{8} = \pm \frac{9}{8} \Rightarrow x = \frac{7}{8} \pm \frac{9}{8} \Rightarrow$

${x}_{1} = \frac{7}{8} - \frac{9}{8} = - \frac{2}{8} = - \frac{1}{4}$

${x}_{2} = \frac{7}{8} + \frac{9}{8} = \frac{16}{8} = 2$.