4*y^2-41=18

Solution for 4*y^2-41=18 equation:

4y^2-41=18

We move all terms to the left:

4y^2-41-(18)=0

We add all the numbers together, and all the variables

4y^2-59=0

a = 4; b = 0; c = -59;
Δ = b2-4ac
Δ = 02-4·4·(-59)
Δ = 944
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{944}=\sqrt{16*59}=\sqrt{16}*\sqrt{59}=4\sqrt{59}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{59}}{2*4}=\frac{0-4\sqrt{59}}{8}
=-\frac{4\sqrt{59}}{8}
=-\frac{\sqrt{59}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{59}}{2*4}=\frac{0+4\sqrt{59}}{8}
=\frac{4\sqrt{59}}{8}
=\frac{\sqrt{59}}{2}
$