10^-40=(10^-8+x)(x)

Solution for 10^-40=(10^-8+x)(x) equation:

10^-40=(10^-8+x)(x)

We move all terms to the left:

10^-40-((10^-8+x)(x))=0

We add all the numbers together, and all the variables

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


We calculate terms in parentheses: -((10^-8+x)x), so:

(10^-8+x)x

We multiply parentheses

10x^2+x^2-8x

We add all the numbers together, and all the variables

11x^2-8x

Back to the equation:

-(11x^2-8x)


We get rid of parentheses

-11x^2+8x=0

a = -11; b = 8; c = 0;
Δ = b2-4ac
Δ = 82-4·(-11)·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8}{2*-11}=\frac{-16}{-22}
=8/11
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8}{2*-11}=\frac{0}{-22}
=0
$