+781 Move the terms around so we have a quadratic equation.
\(2x^2-10x+35=x^2+10\\ x^2-10x-25=0\)
Now we can simply solve for x using the quadratic formula. Let a=1, b=-10, and c=-25.
\(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ x=\frac{10\pm\sqrt{(-10)^2-4(1)(-25)}}{2(1)}\\ x=\frac{10\pm\sqrt{200}}{2}\\ x=\frac{10\pm10\sqrt{2}}{2}\\ x=5\pm5\sqrt{2} \)
So x is \(5+\sqrt{2}\ \text{and}\ 5-\sqrt{2}\).
