1*10^-10=x^2+0.1x

Solution for 1*10^-10=x^2+0.1x equation:

1*10^-10=x^2+0.1x

We move all terms to the left:

1*10^-10-(x^2+0.1x)=0

We add all the numbers together, and all the variables

-(x^2+0.1x)=0

We get rid of parentheses

-x^2-0.1x=0

We add all the numbers together, and all the variables

-1x^2-0.1x=0

a = -1; b = -0.1; c = 0;
Δ = b2-4ac
Δ = -0.12-4·(-1)·0
Δ = 0.01
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-0.1)-\sqrt{0.01}}{2*-1}=\frac{0.1-\sqrt{0.01}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.1)+\sqrt{0.01}}{2*-1}=\frac{0.1+\sqrt{0.01}}{-2}
$