0=x2+19x-42

Solution for 0=x2+19x-42 equation:

0=x2+19x-42

We move all terms to the left:

0-(x2+19x-42)=0

We add all the numbers together, and all the variables

-(+x^2+19x-42)+0=0

We add all the numbers together, and all the variables

-(+x^2+19x-42)=0

We get rid of parentheses

-x^2-19x+42=0

We add all the numbers together, and all the variables

-1x^2-19x+42=0

a = -1; b = -19; c = +42;
Δ = b2-4ac
Δ = -192-4·(-1)·42
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-23}{2*-1}=\frac{-4}{-2}
=+2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+23}{2*-1}=\frac{42}{-2}
=-21
$