0=5v^2+41v+42

Solution for 0=5v^2+41v+42 equation:

0=5v^2+41v+42

We move all terms to the left:

0-(5v^2+41v+42)=0

We add all the numbers together, and all the variables

-(5v^2+41v+42)=0

We get rid of parentheses

-5v^2-41v-42=0

a = -5; b = -41; c = -42;
Δ = b2-4ac
Δ = -412-4·(-5)·(-42)
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-41)-29}{2*-5}=\frac{12}{-10}
=-1+1/5
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-41)+29}{2*-5}=\frac{70}{-10}
=-7
$