0.4u-1.2=2(3/40u+0.4)

Solution for 0.4u-1.2=2(3/40u+0.4) equation:

0.4u-1.2=2(3/40u+0.4)

We move all terms to the left:

0.4u-1.2-(2(3/40u+0.4))=0


Domain of the equation: 40u+0.4))!=0

u∈R


We multiply all the terms by the denominator


(0.4u)*40u-(1.2)*40u+0.4))-(2(3+0.4))=0

We add all the numbers together, and all the variables

(+0.4u)*40u-(1.2)*40u+0.4))-(2(3.4))=0

We add all the numbers together, and all the variables

(+0.4u)*40u-(1.2)*40u=0

We multiply parentheses

0u^2-48u=0

We add all the numbers together, and all the variables

u^2-48u=0

a = 1; b = -48; c = 0;
Δ = b2-4ac
Δ = -482-4·1·0
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-48}{2*1}=\frac{0}{2}
=0
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+48}{2*1}=\frac{96}{2}
=48
$