9n^2+57n=4000

Solution for 9n^2+57n=4000 equation:

9n^2+57n=4000

We move all terms to the left:

9n^2+57n-(4000)=0

a = 9; b = 57; c = -4000;
Δ = b2-4ac
Δ = 572-4·9·(-4000)
Δ = 147249
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{147249}=\sqrt{9*16361}=\sqrt{9}*\sqrt{16361}=3\sqrt{16361}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(57)-3\sqrt{16361}}{2*9}=\frac{-57-3\sqrt{16361}}{18}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(57)+3\sqrt{16361}}{2*9}=\frac{-57+3\sqrt{16361}}{18}
$