(x^2)+(10^-8)x-(10^-14)=0

Solution for (x^2)+(10^-8)x-(10^-14)=0 equation:

(x^2)+(10^-8)x-(10^-14)=0

We add all the numbers together, and all the variables

x^2+(10^-8)x-14-1.0E=0

We multiply parentheses

x^2+10x^2-8x-14-1.0E=0

We add all the numbers together, and all the variables

11x^2-8x-16.718281828459=0

a = 11; b = -8; c = -16.718281828459;
Δ = b2-4ac
Δ = -82-4·11·(-16.718281828459)
Δ = 799.6044004522
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{-(-8)-\sqrt{799.6044004522}}{2*11}=\frac{8-\sqrt{799.6044004522}}{22}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+\sqrt{799.6044004522}}{2*11}=\frac{8+\sqrt{799.6044004522}}{22}
$