a^2+25=64

Solution for a^2+25=64 equation:

a^2+25=64

We move all terms to the left:

a^2+25-(64)=0

We add all the numbers together, and all the variables

a^2-39=0

a = 1; b = 0; c = -39;
Δ = b2-4ac
Δ = 02-4·1·(-39)
Δ = 156
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{156}=\sqrt{4*39}=\sqrt{4}*\sqrt{39}=2\sqrt{39}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{39}}{2*1}=\frac{0-2\sqrt{39}}{2}
=-\frac{2\sqrt{39}}{2}
=-\sqrt{39}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{39}}{2*1}=\frac{0+2\sqrt{39}}{2}
=\frac{2\sqrt{39}}{2}
=\sqrt{39}
$