16y^2+12y^2=396

Solution for 16y^2+12y^2=396 equation:

16y^2+12y^2=396

We move all terms to the left:

16y^2+12y^2-(396)=0

We add all the numbers together, and all the variables

28y^2-396=0

a = 28; b = 0; c = -396;
Δ = b2-4ac
Δ = 02-4·28·(-396)
Δ = 44352
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:

$\sqrt{\Delta}=\sqrt{44352}=\sqrt{576*77}=\sqrt{576}*\sqrt{77}=24\sqrt{77}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24\sqrt{77}}{2*28}=\frac{0-24\sqrt{77}}{56}
=-\frac{24\sqrt{77}}{56}
=-\frac{3\sqrt{77}}{7}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24\sqrt{77}}{2*28}=\frac{0+24\sqrt{77}}{56}
=\frac{24\sqrt{77}}{56}
=\frac{3\sqrt{77}}{7}
$