1.32*10^-5=4x^2+x

Solution for 1.32*10^-5=4x^2+x equation:

1.32*10^-5=4x^2+x

We move all terms to the left:

1.32*10^-5-(4x^2+x)=0

We add all the numbers together, and all the variables

-(4x^2+x)=0

We get rid of parentheses

-4x^2-x=0

We add all the numbers together, and all the variables

-4x^2-1x=0

a = -4; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·(-4)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*-4}=\frac{0}{-8}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*-4}=\frac{2}{-8}
=-1/4
$