(3y+1/4y)-10=39

Solution for (3y+1/4y)-10=39 equation:

(3y+1/4y)-10=39

We move all terms to the left:

(3y+1/4y)-10-(39)=0


Domain of the equation: 4y)!=0

y!=0/1

y!=0

y∈R


We add all the numbers together, and all the variables

(+3y+1/4y)-10-39=0

We add all the numbers together, and all the variables

(+3y+1/4y)-49=0

We get rid of parentheses

3y+1/4y-49=0

We multiply all the terms by the denominator


3y*4y-49*4y+1=0

Wy multiply elements

12y^2-196y+1=0

a = 12; b = -196; c = +1;
Δ = b2-4ac
Δ = -1962-4·12·1
Δ = 38368
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{38368}=\sqrt{16*2398}=\sqrt{16}*\sqrt{2398}=4\sqrt{2398}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-196)-4\sqrt{2398}}{2*12}=\frac{196-4\sqrt{2398}}{24}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-196)+4\sqrt{2398}}{2*12}=\frac{196+4\sqrt{2398}}{24}
$