1/2q+2q+10=-4q-4+1

Solution for 1/2q+2q+10=-4q-4+1 equation:

1/2q+2q+10=-4q-4+1

We move all terms to the left:

1/2q+2q+10-(-4q-4+1)=0


Domain of the equation: 2q!=0

q!=0/2

q!=0

q∈R


We add all the numbers together, and all the variables

1/2q+2q-(-4q-3)+10=0

We add all the numbers together, and all the variables

2q+1/2q-(-4q-3)+10=0

We get rid of parentheses

2q+1/2q+4q+3+10=0

We multiply all the terms by the denominator


2q*2q+4q*2q+3*2q+10*2q+1=0

Wy multiply elements

4q^2+8q^2+6q+20q+1=0

We add all the numbers together, and all the variables

12q^2+26q+1=0

a = 12; b = 26; c = +1;
Δ = b2-4ac
Δ = 262-4·12·1
Δ = 628
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:

$\sqrt{\Delta}=\sqrt{628}=\sqrt{4*157}=\sqrt{4}*\sqrt{157}=2\sqrt{157}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-2\sqrt{157}}{2*12}=\frac{-26-2\sqrt{157}}{24}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+2\sqrt{157}}{2*12}=\frac{-26+2\sqrt{157}}{24}
$