0=x2+26x+48

Solution for 0=x2+26x+48 equation:

0=x2+26x+48

We move all terms to the left:

0-(x2+26x+48)=0

We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

-x^2-26x-48=0

We add all the numbers together, and all the variables

-1x^2-26x-48=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
$