x^2+48=26x

Solution for x^2+48=26x equation:

x^2+48=26x

We move all terms to the left:

x^2+48-(26x)=0

a = 1; b = -26; c = +48;
Δ = b2-4ac
Δ = -262-4·1·48
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-22}{2*1}=\frac{4}{2}
=2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+22}{2*1}=\frac{48}{2}
=24
$