The quadratic is
[tex]x^2+35=5x[/tex]Subtract 5x from both sides
[tex]\begin{gathered} x^2-5x+35=5x-5x \\ x^2-5x+35=0 \end{gathered}[/tex]To solve it we will use the formula
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]a is the coefficient of x^2
b is the coefficient of x
c is the numerical term
Since a = 1, b = -5, c = 35
Then substitute them in the formula above
[tex]x=\frac{-(-5)\pm\sqrt[]{(-5)^2-4(1)(35)}}{2(1)}[/tex]Simplify it
[tex]\begin{gathered} x=\frac{5\pm\sqrt[]{25-140}}{2} \\ x=\frac{5\pm\sqrt[]{-115}}{2} \end{gathered}[/tex]since the square root of a -ve number is imaginary, then we will use i
[tex]\begin{gathered} x=\frac{5\pm\sqrt[]{115}.\sqrt[]{-1}}{2} \\ x=\frac{5\pm\sqrt[]{115}i}{2} \\ \sqrt[]{-1}=i \end{gathered}[/tex]The solutions are
[tex]\mleft\lbrace\frac{5+\sqrt[]{115}i}{2},\frac{5-\sqrt[]{115}i}{2}\mright\rbrace[/tex]