0.4x+10=3/4x+12

Solution for 0.4x+10=3/4x+12 equation:

0.4x+10=3/4x+12

We move all terms to the left:

0.4x+10-(3/4x+12)=0


Domain of the equation: 4x+12)!=0

x∈R


We get rid of parentheses

0.4x-3/4x-12+10=0

We multiply all the terms by the denominator


(0.4x)*4x-12*4x+10*4x-3=0

We add all the numbers together, and all the variables

(+0.4x)*4x-12*4x+10*4x-3=0

We multiply parentheses

0x^2-12*4x+10*4x-3=0

Wy multiply elements

0x^2-48x+40x-3=0

We add all the numbers together, and all the variables

x^2-8x-3=0

a = 1; b = -8; c = -3;
Δ = b2-4ac
Δ = -82-4·1·(-3)
Δ = 76
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:

$\sqrt{\Delta}=\sqrt{76}=\sqrt{4*19}=\sqrt{4}*\sqrt{19}=2\sqrt{19}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{19}}{2*1}=\frac{8-2\sqrt{19}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{19}}{2*1}=\frac{8+2\sqrt{19}}{2}
$