we have[tex]0=5x^{2} -2x+6[/tex]Group terms that contain the same variable, and move the constant to the opposite side of the equation[tex]5x^{2} -2x=-6[/tex]Factor the leading coefficient [tex]5(x^{2} -(2/5)x)=-6[/tex]Complete the square. Remember to balance the equation by adding the same constants to each side[tex]5(x^{2} -(2/5)x+(1/25))=-6+(1/5)[/tex] [tex]5(x-(1/5))^{2}=-(29/5)[/tex] [tex](x-(1/5))^{2}=-(29/25)[/tex] remember that[tex]i=\sqrt{-1}[/tex][tex](x-(1/5))^{2}=-(29/25)\\ \\ (x-(1/5))=(+/-)\frac{\sqrt{29}} {5}i\\ \\ x1=\frac{1}{5} +\frac{\sqrt{29}} {5}i\\ \\ x2=\frac{1}{5} -\frac{\sqrt{29}} {5}i[/tex]thereforethe answer isthe solutions of the quadratic equation are[tex]x1=\frac{1}{5} +\frac{\sqrt{29}} {5}i[/tex][tex]x2=\frac{1}{5} -\frac{\sqrt{29}} {5}i[/tex]