7/2q+5=-3+2/q

Solution for 7/2q+5=-3+2/q equation:

7/2q+5=-3+2/q

We move all terms to the left:

7/2q+5-(-3+2/q)=0


Domain of the equation: 2q!=0

q!=0/2

q!=0

q∈R



Domain of the equation: q)!=0

q!=0/1

q!=0

q∈R


We add all the numbers together, and all the variables

7/2q-(2/q-3)+5=0

We get rid of parentheses

7/2q-2/q+3+5=0

We calculate fractions


7q/2q^2+(-4q)/2q^2+3+5=0

We add all the numbers together, and all the variables

7q/2q^2+(-4q)/2q^2+8=0

We multiply all the terms by the denominator


7q+(-4q)+8*2q^2=0

Wy multiply elements

16q^2+7q+(-4q)=0

We get rid of parentheses

16q^2+7q-4q=0

We add all the numbers together, and all the variables

16q^2+3q=0

a = 16; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·16·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*16}=\frac{-6}{32}
=-3/16
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*16}=\frac{0}{32}
=0
$