Move the constant on the right side to the left side by subtracting both sides by 35
[tex]\begin{gathered} x^2-10x+25=35 \\ x^2-10x+25-35=35-35 \\ x^2-10x-10=\cancel{0} \end{gathered}[/tex]Now that it is in the standard form of quadratic equation, we can use the quadratic formula to solve for x
[tex]\begin{gathered} \text{The standard form is }ax^2+bx+c=0 \\ \\ \text{In }x^2-10x-10=0, \\ a=1,b=-10,c=-10 \end{gathered}[/tex][tex]\begin{gathered} \text{Use the quadratic formula} \\ x=\frac{ -b \pm\sqrt{b^2 - 4ac}}{ 2a } \\ \\ \text{Substitute the values for }a,b\text{ and }c \\ x=\frac{ -(-10) \pm\sqrt{(-10)^2 - 4(1)(-10)}}{ 2(1) } \\ x=\frac{10\pm\sqrt[]{100-(-40)}}{2} \\ x=\frac{10\pm\sqrt[]{100+40}}{2} \\ x=\frac{ 10 \pm\sqrt{140}}{ 2 } \\ x=\frac{ 10 \pm2\sqrt{35}\, }{ 2 } \\ x=\frac{ 10 }{ 2 }\pm\frac{2\sqrt{35}\, }{ 2 } \\ x=5\pm\frac{\cancel{2}\sqrt[]{35}}{\cancel{2}} \\ x=5\pm\sqrt[]{35} \end{gathered}[/tex]Therefore the solution for the given equation is
[tex]\begin{gathered} x=5+\sqrt[]{35}\text{ or }x=10.9161 \\ \text{AND} \\ x=5-\sqrt[]{35}\text{ or }x=-0.9161 \end{gathered}[/tex]