2n(4n-19)=-35

Solution for 2n(4n-19)=-35 equation:

2n(4n-19)=-35

We move all terms to the left:

2n(4n-19)-(-35)=0

We add all the numbers together, and all the variables

2n(4n-19)+35=0

We multiply parentheses

8n^2-38n+35=0

a = 8; b = -38; c = +35;
Δ = b2-4ac
Δ = -382-4·8·35
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-38)-18}{2*8}=\frac{20}{16}
=1+1/4
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-38)+18}{2*8}=\frac{56}{16}
=3+1/2
$