Answer:
[tex]$x_{1}=-\frac{2+\sqrt{34}}{6}$[/tex]
[tex]$x_{2}=-\frac{2-\sqrt{34}}{6}$[/tex]
Step-by-step explanation:
Quadratic Equation given:
[tex]6x^2 + 4x - 5 = 0[/tex]
Start dividing both sides by 6.
[tex]$\frac{6}{6} x^2 +\frac{4}{6} x - \frac{5}{6} = \frac{0}{6} $[/tex]
[tex]$x^2 +\frac{2}{3} x - \frac{5}{6} = 0 $[/tex]
[tex]$x^2 +\frac{2}{3} x = \frac{5}{6} $[/tex]
Once, [tex]$\left( \frac{2}{3} \cdot \frac{1}{2} \right)^2=\frac{1}{9} $[/tex]
[tex]$x^2 +\frac{2}{3} x+\frac{1}{9} = \frac{5}{6}+\frac{1}{9} $[/tex]
[tex]$\left(x+\frac{1}{3} \right)^2 = \frac{17}{18}} $[/tex]
[tex]$x+\frac{1}{3} =\pm\sqrt{\frac{17}{18}} } $[/tex]
Solving [tex]$\sqrt{\frac{17}{18}} $[/tex]
[tex]$\sqrt{\frac{17(18)}{18(18)}}= \sqrt{\frac{306}{324}}=\frac{3\sqrt{34} }{18} =\frac{\sqrt{34} }{6} $[/tex]
Then,
[tex]$x+\frac{1}{3} =\pm\frac{\sqrt{34} }{6} $[/tex]
[tex]$x =-\frac{1}{3} \pm\frac{\sqrt{34} }{6} $[/tex]
[tex]$x_{1}=-\frac{2+\sqrt{34}}{6}$[/tex]
[tex]$x_{2}=-\frac{2-\sqrt{34}}{6}$[/tex]
