1/4y2+20=36

Solution for 1/4y2+20=36 equation:

1/4y^2+20=36

We move all terms to the left:

1/4y^2+20-(36)=0


Domain of the equation: 4y^2!=0

y^2!=0/4

y^2!=√0

y!=0

y∈R


We add all the numbers together, and all the variables

1/4y^2-16=0

We multiply all the terms by the denominator


-16*4y^2+1=0

Wy multiply elements

-64y^2+1=0

a = -64; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-64)·1
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16}{2*-64}=\frac{-16}{-128}
=1/8
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16}{2*-64}=\frac{16}{-128}
=-1/8
$