16n^2+40n+25=4

Solution for 16n^2+40n+25=4 equation:

16n^2+40n+25=4

We move all terms to the left:

16n^2+40n+25-(4)=0

We add all the numbers together, and all the variables

16n^2+40n+21=0

a = 16; b = 40; c = +21;
Δ = b2-4ac
Δ = 402-4·16·21
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-16}{2*16}=\frac{-56}{32}
=-1+3/4
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+16}{2*16}=\frac{-24}{32}
=-3/4
$