x^2=10x+25=35

Solution for x^2=10x+25=35 equation:

x^2=10x+25=35

We move all terms to the left:

x^2-(10x+25)=0

We get rid of parentheses

x^2-10x-25=0

a = 1; b = -10; c = -25;
Δ = b2-4ac
Δ = -102-4·1·(-25)
Δ = 200
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{200}=\sqrt{100*2}=\sqrt{100}*\sqrt{2}=10\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-10\sqrt{2}}{2*1}=\frac{10-10\sqrt{2}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+10\sqrt{2}}{2*1}=\frac{10+10\sqrt{2}}{2}
$