4n^2=28n

Solution for 4n^2=28n equation:

4n^2=28n

We move all terms to the left:

4n^2-(28n)=0

a = 4; b = -28; c = 0;
Δ = b2-4ac
Δ = -282-4·4·0
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{-(-28)-28}{2*4}=\frac{0}{8}
=0
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+28}{2*4}=\frac{56}{8}
=7
$