y^2+y^2-6y-3-6y=14

Solution for y^2+y^2-6y-3-6y=14 equation:

y^2+y^2-6y-3-6y=14

We move all terms to the left:

y^2+y^2-6y-3-6y-(14)=0

We add all the numbers together, and all the variables

2y^2-12y-17=0

a = 2; b = -12; c = -17;
Δ = b2-4ac
Δ = -122-4·2·(-17)
Δ = 280
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:

$\sqrt{\Delta}=\sqrt{280}=\sqrt{4*70}=\sqrt{4}*\sqrt{70}=2\sqrt{70}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{70}}{2*2}=\frac{12-2\sqrt{70}}{4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{70}}{2*2}=\frac{12+2\sqrt{70}}{4}
$