0.13x+0.3x(x-1)=0.01)(3x-4)

Solution for 0.13x+0.3x(x-1)=0.01)(3x-4) equation:

0.13x+0.3x(x-1)=0.01)(3x-4)

We move all terms to the left:

0.13x+0.3x(x-1)-(0.01)(3x-4))=0

We multiply parentheses

0x^2+0.13x+0x-(0.01)(3x-4))=0

We multiply parentheses ..

0x^2+0.13x+0x-(+0.03x-0.04))=0

We add all the numbers together, and all the variables

0x^2+0.13x+0x-(0.03x-0.04))=0

We add all the numbers together, and all the variables

x^2+1.13x-(0.03x-0.04))=0

We get rid of parentheses

x^2+1.13x-0.03x+0.04)=0

We add all the numbers together, and all the variables

x^2+1.1x=0

a = 1; b = 1.1; c = 0;
Δ = b2-4ac
Δ = 1.12-4·1·0
Δ = 1.21
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{-(1.1)-\sqrt{1.21}}{2*1}=\frac{-1.1-\sqrt{1.21}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.1)+\sqrt{1.21}}{2*1}=\frac{-1.1+\sqrt{1.21}}{2}
$