EXPLANATION
Since we have the function 5x^2-5x+6=0
[tex]\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}[/tex][tex]x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex][tex]\mathrm{For\:}\quad a=5,\:b=-5,\:c=6[/tex][tex]x_{1,\:2}=\frac{-\left(-5\right)\pm \sqrt{\left(-5\right)^2-4\cdot \:5\cdot \:6}}{2\cdot \:5}[/tex]Computing the powers:
[tex]=\sqrt{5^2-4\cdot \:5\cdot \:6}[/tex][tex]\mathrm{Multiply\:the\:numbers:}\:4\cdot \:5\cdot \:6=120[/tex][tex]=\sqrt{5^2-120}[/tex]Apply imaginary number rule:
[tex]=\sqrt{5^2-120}[/tex][tex]=\sqrt{95}i[/tex][tex]x_{1,\:2}=\frac{-\left(-5\right)\pm \sqrt{95}i}{2\cdot \:5}[/tex][tex]Separate\:the\:solutions[/tex][tex]x_1=\frac{-\left(-5\right)+\sqrt{95}i}{2\cdot \:5},\:x_2=\frac{-\left(-5\right)-\sqrt{95}i}{2\cdot \:5}[/tex]Simplifying:
[tex]\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}[/tex][tex]x=\frac{1}{2}+i\frac{\sqrt{95}}{10},\:x=\frac{1}{2}-i\frac{\sqrt{95}}{10}[/tex]In conclusion, the function have two complex solutions.
