3w-13=(1/4)(52-12w)

Solution for 3w-13=(1/4)(52-12w) equation:

3w-13=(1/4)(52-12w)

We move all terms to the left:

3w-13-((1/4)(52-12w))=0


Domain of the equation: 4)(52-12w))!=0

w∈R


We add all the numbers together, and all the variables

3w-((+1/4)(-12w+52))-13=0

We multiply parentheses ..

-((-12w^2+1/4*52))+3w-13=0

We multiply all the terms by the denominator


-((-12w^2+1+3w*4*52))-13*4*52))=0


We calculate terms in parentheses: -((-12w^2+1+3w*4*52)), so:

(-12w^2+1+3w*4*52)

We get rid of parentheses

-12w^2+3w*4*52+1

Wy multiply elements

-12w^2+624w*5+1

Wy multiply elements

-12w^2+3120w+1

Back to the equation:

-(-12w^2+3120w+1)


We add all the numbers together, and all the variables

-(-12w^2+3120w+1)=0

We get rid of parentheses

12w^2-3120w-1=0

a = 12; b = -3120; c = -1;
Δ = b2-4ac
Δ = -31202-4·12·(-1)
Δ = 9734448
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{9734448}=\sqrt{16*608403}=\sqrt{16}*\sqrt{608403}=4\sqrt{608403}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3120)-4\sqrt{608403}}{2*12}=\frac{3120-4\sqrt{608403}}{24}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3120)+4\sqrt{608403}}{2*12}=\frac{3120+4\sqrt{608403}}{24}
$