25q+q=1/2q-3

Solution for 25q+q=1/2q-3 equation:

25q+q=1/2q-3

We move all terms to the left:

25q+q-(1/2q-3)=0


Domain of the equation: 2q-3)!=0

q∈R


We add all the numbers together, and all the variables

26q-(1/2q-3)=0

We get rid of parentheses

26q-1/2q+3=0

We multiply all the terms by the denominator


26q*2q+3*2q-1=0

Wy multiply elements

52q^2+6q-1=0

a = 52; b = 6; c = -1;
Δ = b2-4ac
Δ = 62-4·52·(-1)
Δ = 244
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:

$\sqrt{\Delta}=\sqrt{244}=\sqrt{4*61}=\sqrt{4}*\sqrt{61}=2\sqrt{61}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{61}}{2*52}=\frac{-6-2\sqrt{61}}{104}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{61}}{2*52}=\frac{-6+2\sqrt{61}}{104}
$