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]
therefore
the answer is
the 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]
