y^2+8y-64=3

Solution for y^2+8y-64=3 equation:

y^2+8y-64=3

We move all terms to the left:

y^2+8y-64-(3)=0

We add all the numbers together, and all the variables

y^2+8y-67=0

a = 1; b = 8; c = -67;
Δ = b2-4ac
Δ = 82-4·1·(-67)
Δ = 332
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:

$\sqrt{\Delta}=\sqrt{332}=\sqrt{4*83}=\sqrt{4}*\sqrt{83}=2\sqrt{83}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{83}}{2*1}=\frac{-8-2\sqrt{83}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{83}}{2*1}=\frac{-8+2\sqrt{83}}{2}
$