2y^2+14=110

Solution for 2y^2+14=110 equation:

2y^2+14=110

We move all terms to the left:

2y^2+14-(110)=0

We add all the numbers together, and all the variables

2y^2-96=0

a = 2; b = 0; c = -96;
Δ = b2-4ac
Δ = 02-4·2·(-96)
Δ = 768
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:

$\sqrt{\Delta}=\sqrt{768}=\sqrt{256*3}=\sqrt{256}*\sqrt{3}=16\sqrt{3}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{3}}{2*2}=\frac{0-16\sqrt{3}}{4}
=-\frac{16\sqrt{3}}{4}
=-4\sqrt{3}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{3}}{2*2}=\frac{0+16\sqrt{3}}{4}
=\frac{16\sqrt{3}}{4}
=4\sqrt{3}
$