(4)(n+2)(n+5)=280

Solution for (4)(n+2)(n+5)=280 equation:

(4)(n+2)(n+5)=280

We move all terms to the left:

(4)(n+2)(n+5)-(280)=0

We multiply parentheses ..

4(+n^2+5n+2n+10)-280=0

We multiply parentheses

4n^2+20n+8n+40-280=0

We add all the numbers together, and all the variables

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
$