0=x2-19x+48

Solution for 0=x2-19x+48 equation:

0=x2-19x+48

We move all terms to the left:

0-(x2-19x+48)=0

We add all the numbers together, and all the variables

-(+x^2-19x+48)+0=0

We add all the numbers together, and all the variables

-(+x^2-19x+48)=0

We get rid of parentheses

-x^2+19x-48=0

We add all the numbers together, and all the variables

-1x^2+19x-48=0

a = -1; b = 19; c = -48;
Δ = b2-4ac
Δ = 192-4·(-1)·(-48)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-13}{2*-1}=\frac{-32}{-2}
=+16
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+13}{2*-1}=\frac{-6}{-2}
=+3
$