-2(q-29)=(1/2)(4-8q)

Solution for -2(q-29)=(1/2)(4-8q) equation:

-2(q-29)=(1/2)(4-8q)

We move all terms to the left:

-2(q-29)-((1/2)(4-8q))=0


Domain of the equation: 2)(4-8q))!=0

q∈R


We add all the numbers together, and all the variables

-2(q-29)-((+1/2)(-8q+4))=0

We multiply parentheses

-2q-((+1/2)(-8q+4))+58=0

We multiply parentheses ..

-((-8q^2+1/2*4))-2q+58=0

We multiply all the terms by the denominator


-((-8q^2+1-2q*2*4))+58*2*4))=0


We calculate terms in parentheses: -((-8q^2+1-2q*2*4)), so:

(-8q^2+1-2q*2*4)

We get rid of parentheses

-8q^2-2q*2*4+1

Wy multiply elements

-8q^2-16q*4+1

Wy multiply elements

-8q^2-64q+1

Back to the equation:

-(-8q^2-64q+1)


We add all the numbers together, and all the variables

-(-8q^2-64q+1)=0

We get rid of parentheses

8q^2+64q-1=0

a = 8; b = 64; c = -1;
Δ = b2-4ac
Δ = 642-4·8·(-1)
Δ = 4128
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{4128}=\sqrt{16*258}=\sqrt{16}*\sqrt{258}=4\sqrt{258}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(64)-4\sqrt{258}}{2*8}=\frac{-64-4\sqrt{258}}{16}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(64)+4\sqrt{258}}{2*8}=\frac{-64+4\sqrt{258}}{16}
$