4n^2-22=27

Solution for 4n^2-22=27 equation:

4n^2-22=27

We move all terms to the left:

4n^2-22-(27)=0

We add all the numbers together, and all the variables

4n^2-49=0

a = 4; b = 0; c = -49;
Δ = b2-4ac
Δ = 02-4·4·(-49)
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-28}{2*4}=\frac{-28}{8}
=-3+1/2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+28}{2*4}=\frac{28}{8}
=3+1/2
$