1000=x^2+10x

Solution for 1000=x^2+10x equation:

1000=x^2+10x

We move all terms to the left:

1000-(x^2+10x)=0

We get rid of parentheses

-x^2-10x+1000=0

We add all the numbers together, and all the variables

-1x^2-10x+1000=0

a = -1; b = -10; c = +1000;
Δ = b2-4ac
Δ = -102-4·(-1)·1000
Δ = 4100
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:

$\sqrt{\Delta}=\sqrt{4100}=\sqrt{100*41}=\sqrt{100}*\sqrt{41}=10\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-10\sqrt{41}}{2*-1}=\frac{10-10\sqrt{41}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+10\sqrt{41}}{2*-1}=\frac{10+10\sqrt{41}}{-2}
$