g^2+26g+25=0

Solution for g^2+26g+25=0 equation:

g^2+26g+25=0

a = 1; b = 26; c = +25;
Δ = b2-4ac
Δ = 262-4·1·25
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-24}{2*1}=\frac{-50}{2}
=-25
$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+24}{2*1}=\frac{-2}{2}
=-1
$