y^2+6=17

Solution for y^2+6=17 equation:

y^2+6=17

We move all terms to the left:

y^2+6-(17)=0

We add all the numbers together, and all the variables

y^2-11=0

a = 1; b = 0; c = -11;
Δ = b2-4ac
Δ = 02-4·1·(-11)
Δ = 44
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:

$\sqrt{\Delta}=\sqrt{44}=\sqrt{4*11}=\sqrt{4}*\sqrt{11}=2\sqrt{11}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{11}}{2*1}=\frac{0-2\sqrt{11}}{2}
=-\frac{2\sqrt{11}}{2}
=-\sqrt{11}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{11}}{2*1}=\frac{0+2\sqrt{11}}{2}
=\frac{2\sqrt{11}}{2}
=\sqrt{11}
$