4n2-7=27n

Solution for 4n2-7=27n equation:

4n^2-7=27n

We move all terms to the left:

4n^2-7-(27n)=0

a = 4; b = -27; c = -7;
Δ = b2-4ac
Δ = -272-4·4·(-7)
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-27)-29}{2*4}=\frac{-2}{8}
=-1/4
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+29}{2*4}=\frac{56}{8}
=7
$