25+q=1/2q=3

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

25+q=1/2q=3

We move all terms to the left:

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


Domain of the equation: 2q)!=0

q!=0/1

q!=0

q∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

q-1/2q+25=0

We multiply all the terms by the denominator


q*2q+25*2q-1=0

Wy multiply elements

2q^2+50q-1=0

a = 2; b = 50; c = -1;
Δ = b2-4ac
Δ = 502-4·2·(-1)
Δ = 2508
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:

$\sqrt{\Delta}=\sqrt{2508}=\sqrt{4*627}=\sqrt{4}*\sqrt{627}=2\sqrt{627}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-2\sqrt{627}}{2*2}=\frac{-50-2\sqrt{627}}{4}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+2\sqrt{627}}{2*2}=\frac{-50+2\sqrt{627}}{4}
$