5=1/4(q^2)

Solution for 5=1/4(q^2) equation:

5=1/4(q^2)

We move all terms to the left:

5-(1/4(q^2))=0


Domain of the equation: 4q^2)!=0

q!=0/1

q!=0

q∈R


We get rid of parentheses

-1/4q^2+5=0

We multiply all the terms by the denominator


5*4q^2-1=0

Wy multiply elements

20q^2-1=0

a = 20; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·20·(-1)
Δ = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{5}}{2*20}=\frac{0-4\sqrt{5}}{40}
=-\frac{4\sqrt{5}}{40}
=-\frac{\sqrt{5}}{10}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{5}}{2*20}=\frac{0+4\sqrt{5}}{40}
=\frac{4\sqrt{5}}{40}
=\frac{\sqrt{5}}{10}
$