2v^2-19v+42=0

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

2v^2-19v+42=0

a = 2; b = -19; c = +42;
Δ = b2-4ac
Δ = -192-4·2·42
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-5}{2*2}=\frac{14}{4}
=3+1/2
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+5}{2*2}=\frac{24}{4}
=6
$