40n2+2n+4=3n+10

Solution for 40n2+2n+4=3n+10 equation:

40n^2+2n+4=3n+10

We move all terms to the left:

40n^2+2n+4-(3n+10)=0

We get rid of parentheses

40n^2+2n-3n-10+4=0

We add all the numbers together, and all the variables

40n^2-1n-6=0

a = 40; b = -1; c = -6;
Δ = b2-4ac
Δ = -12-4·40·(-6)
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-31}{2*40}=\frac{-30}{80}
=-3/8
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+31}{2*40}=\frac{32}{80}
=2/5
$