2n^2=-20n-42

Solution for 2n^2=-20n-42 equation:

2n^2=-20n-42

We move all terms to the left:

2n^2-(-20n-42)=0

We get rid of parentheses

2n^2+20n+42=0

a = 2; b = 20; c = +42;
Δ = b2-4ac
Δ = 202-4·2·42
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-8}{2*2}=\frac{-28}{4}
=-7
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+8}{2*2}=\frac{-12}{4}
=-3
$