2y^2=58+16y

Solution for 2y^2=58+16y equation:

2y^2=58+16y

We move all terms to the left:

2y^2-(58+16y)=0

We add all the numbers together, and all the variables

2y^2-(16y+58)=0

We get rid of parentheses

2y^2-16y-58=0

a = 2; b = -16; c = -58;
Δ = b2-4ac
Δ = -162-4·2·(-58)
Δ = 720
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:

$\sqrt{\Delta}=\sqrt{720}=\sqrt{144*5}=\sqrt{144}*\sqrt{5}=12\sqrt{5}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-12\sqrt{5}}{2*2}=\frac{16-12\sqrt{5}}{4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+12\sqrt{5}}{2*2}=\frac{16+12\sqrt{5}}{4}
$