4n^2+22n=12

Solution for 4n^2+22n=12 equation:

4n^2+22n=12

We move all terms to the left:

4n^2+22n-(12)=0

a = 4; b = 22; c = -12;
Δ = b2-4ac
Δ = 222-4·4·(-12)
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-26}{2*4}=\frac{-48}{8}
=-6
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+26}{2*4}=\frac{4}{8}
=1/2
$