10=3+6y^2

Solution for 10=3+6y^2 equation:

10=3+6y^2

We move all terms to the left:

10-(3+6y^2)=0

We get rid of parentheses

-6y^2-3+10=0

We add all the numbers together, and all the variables

-6y^2+7=0

a = -6; b = 0; c = +7;
Δ = b2-4ac
Δ = 02-4·(-6)·7
Δ = 168
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:

$\sqrt{\Delta}=\sqrt{168}=\sqrt{4*42}=\sqrt{4}*\sqrt{42}=2\sqrt{42}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{42}}{2*-6}=\frac{0-2\sqrt{42}}{-12}
=-\frac{2\sqrt{42}}{-12}
=-\frac{\sqrt{42}}{-6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{42}}{2*-6}=\frac{0+2\sqrt{42}}{-12}
=\frac{2\sqrt{42}}{-12}
=\frac{\sqrt{42}}{-6}
$