-42-19x+x^2=0

Solution for -42-19x+x^2=0 equation:

-42-19x+x^2=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
$