0.4x+10=3/5x+12

Solution for 0.4x+10=3/5x+12 equation:

0.4x+10=3/5x+12

We move all terms to the left:

0.4x+10-(3/5x+12)=0


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

x∈R


We get rid of parentheses

0.4x-3/5x-12+10=0

We multiply all the terms by the denominator


(0.4x)*5x-12*5x+10*5x-3=0

We add all the numbers together, and all the variables

(+0.4x)*5x-12*5x+10*5x-3=0

We multiply parentheses

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

Wy multiply elements

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

We add all the numbers together, and all the variables

x^2-10x-3=0

a = 1; b = -10; c = -3;
Δ = b2-4ac
Δ = -102-4·1·(-3)
Δ = 112
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:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-4\sqrt{7}}{2*1}=\frac{10-4\sqrt{7}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+4\sqrt{7}}{2*1}=\frac{10+4\sqrt{7}}{2}
$