4y^2-13=49

Solution for 4y^2-13=49 equation:

4y^2-13=49

We move all terms to the left:

4y^2-13-(49)=0

We add all the numbers together, and all the variables

4y^2-62=0

a = 4; b = 0; c = -62;
Δ = b2-4ac
Δ = 02-4·4·(-62)
Δ = 992
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:

$\sqrt{\Delta}=\sqrt{992}=\sqrt{16*62}=\sqrt{16}*\sqrt{62}=4\sqrt{62}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{62}}{2*4}=\frac{0-4\sqrt{62}}{8}
=-\frac{4\sqrt{62}}{8}
=-\frac{\sqrt{62}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{62}}{2*4}=\frac{0+4\sqrt{62}}{8}
=\frac{4\sqrt{62}}{8}
=\frac{\sqrt{62}}{2}
$