10n^2+27n+9=4

Solution for 10n^2+27n+9=4 equation:

10n^2+27n+9=4

We move all terms to the left:

10n^2+27n+9-(4)=0

We add all the numbers together, and all the variables

10n^2+27n+5=0

a = 10; b = 27; c = +5;
Δ = b2-4ac
Δ = 272-4·10·5
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-23}{2*10}=\frac{-50}{20}
=-2+1/2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+23}{2*10}=\frac{-4}{20}
=-1/5
$