10x+35=x^2+40

Solution for 10x+35=x^2+40 equation:

10x+35=x^2+40

We move all terms to the left:

10x+35-(x^2+40)=0

We get rid of parentheses

-x^2+10x-40+35=0

We add all the numbers together, and all the variables

-1x^2+10x-5=0

a = -1; b = 10; c = -5;
Δ = b2-4ac
Δ = 102-4·(-1)·(-5)
Δ = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-4\sqrt{5}}{2*-1}=\frac{-10-4\sqrt{5}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+4\sqrt{5}}{2*-1}=\frac{-10+4\sqrt{5}}{-2}
$