.4x+10=3/5x+12

Solution for .4x+10=3/5x+12 equation:

.4x+10=3/5x+12

We move all terms to the left:

.4x+10-(3/5x+12)=0


Domain of the equation: 5x+12)!=0

x∈R


We get rid of parentheses

.4x-3/5x-12+10=0

We multiply all the terms by the denominator


(.4x)*5x-12*5x+10*5x-3=0

We add all the numbers together, and all the variables

(+.4x)*5x-12*5x+10*5x-3=0

We multiply parentheses

5x^2-12*5x+10*5x-3=0

Wy multiply elements

5x^2-60x+50x-3=0

We add all the numbers together, and all the variables

5x^2-10x-3=0

a = 5; b = -10; c = -3;
Δ = b2-4ac
Δ = -102-4·5·(-3)
Δ = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-4\sqrt{10}}{2*5}=\frac{10-4\sqrt{10}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+4\sqrt{10}}{2*5}=\frac{10+4\sqrt{10}}{10}
$