x2+19x-42=0

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

x2+19x-42=0

We add all the numbers together, and all the variables

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