4n^2+28n=240

Solution for 4n^2+28n=240 equation:

4n^2+28n=240

We move all terms to the left:

4n^2+28n-(240)=0

a = 4; b = 28; c = -240;
Δ = b2-4ac
Δ = 282-4·4·(-240)
Δ = 4624
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4624}=68$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-68}{2*4}=\frac{-96}{8}
=-12
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+68}{2*4}=\frac{40}{8}
=5
$