1000(1+x^2)=1000+440

Solution for 1000(1+x^2)=1000+440 equation:

1000(1+x^2)=1000+440

We move all terms to the left:

1000(1+x^2)-(1000+440)=0

We add all the numbers together, and all the variables

1000(1+x^2)-1440=0

We multiply parentheses

1000x^2+1000-1440=0

We add all the numbers together, and all the variables

1000x^2-440=0

a = 1000; b = 0; c = -440;
Δ = b2-4ac
Δ = 02-4·1000·(-440)
Δ = 1760000
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:

$\sqrt{\Delta}=\sqrt{1760000}=\sqrt{160000*11}=\sqrt{160000}*\sqrt{11}=400\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-400\sqrt{11}}{2*1000}=\frac{0-400\sqrt{11}}{2000}
=-\frac{400\sqrt{11}}{2000}
=-\frac{\sqrt{11}}{5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+400\sqrt{11}}{2*1000}=\frac{0+400\sqrt{11}}{2000}
=\frac{400\sqrt{11}}{2000}
=\frac{\sqrt{11}}{5}
$