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

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

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

We move all terms to the left:

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


Domain of the equation: 8)q!=0

q!=0/1

q!=0

q∈R


We add all the numbers together, and all the variables

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

We multiply parentheses

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

We get rid of parentheses

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

We calculate fractions


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


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

-((q-1)*2)/()+()/(

We add all the numbers together, and all the variables

-((q-1)*2)/()+1

We multiply all the terms by the denominator


-((q-1)*2)+1*()


We calculate terms in parentheses: -((q-1)*2), so:

(q-1)*2

We multiply parentheses

2q-2

Back to the equation:

-(2q-2)


We add all the numbers together, and all the variables

-(2q-2)

We get rid of parentheses

-2q+2

Back to the equation:

+(-2q+2)


We add all the numbers together, and all the variables

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

We get rid of parentheses

-1q^2-2q+2=0

a = -1; b = -2; c = +2;
Δ = b2-4ac
Δ = -22-4·(-1)·2
Δ = 12
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:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{3}}{2*-1}=\frac{2-2\sqrt{3}}{-2}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{3}}{2*-1}=\frac{2+2\sqrt{3}}{-2}
$