8y^2+12*(2^.5)y=306

Solution for 8y^2+12*(2^.5)y=306 equation:

8y^2+12(2^.5)y=306

We move all terms to the left:

8y^2+12(2^.5)y-(306)=0

We multiply parentheses

8y^2+24y^2-306=0

We add all the numbers together, and all the variables

32y^2-306=0

a = 32; b = 0; c = -306;
Δ = b2-4ac
Δ = 02-4·32·(-306)
Δ = 39168
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:

$\sqrt{\Delta}=\sqrt{39168}=\sqrt{2304*17}=\sqrt{2304}*\sqrt{17}=48\sqrt{17}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-48\sqrt{17}}{2*32}=\frac{0-48\sqrt{17}}{64}
=-\frac{48\sqrt{17}}{64}
=-\frac{3\sqrt{17}}{4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+48\sqrt{17}}{2*32}=\frac{0+48\sqrt{17}}{64}
=\frac{48\sqrt{17}}{64}
=\frac{3\sqrt{17}}{4}
$