0.4x+6=2/3x+10

Solution for 0.4x+6=2/3x+10 equation:

0.4x+6=2/3x+10

We move all terms to the left:

0.4x+6-(2/3x+10)=0


Domain of the equation: 3x+10)!=0

x∈R


We get rid of parentheses

0.4x-2/3x-10+6=0

We multiply all the terms by the denominator


(0.4x)*3x-10*3x+6*3x-2=0

We add all the numbers together, and all the variables

(+0.4x)*3x-10*3x+6*3x-2=0

We multiply parentheses

0x^2-10*3x+6*3x-2=0

Wy multiply elements

0x^2-30x+18x-2=0

We add all the numbers together, and all the variables

x^2-12x-2=0

a = 1; b = -12; c = -2;
Δ = b2-4ac
Δ = -122-4·1·(-2)
Δ = 152
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{152}=\sqrt{4*38}=\sqrt{4}*\sqrt{38}=2\sqrt{38}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{38}}{2*1}=\frac{12-2\sqrt{38}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{38}}{2*1}=\frac{12+2\sqrt{38}}{2}
$