0=-4(n^2-25n+45)

Solution for 0=-4(n^2-25n+45) equation:

0=-4(n^2-25n+45)

We move all terms to the left:

0-(-4(n^2-25n+45))=0

We add all the numbers together, and all the variables

-(-4(n^2-25n+45))=0


We calculate terms in parentheses: -(-4(n^2-25n+45)), so:

-4(n^2-25n+45)

We multiply parentheses

-4n^2+100n-180

Back to the equation:

-(-4n^2+100n-180)


We get rid of parentheses

4n^2-100n+180=0

a = 4; b = -100; c = +180;
Δ = b2-4ac
Δ = -1002-4·4·180
Δ = 7120
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:

$\sqrt{\Delta}=\sqrt{7120}=\sqrt{16*445}=\sqrt{16}*\sqrt{445}=4\sqrt{445}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-100)-4\sqrt{445}}{2*4}=\frac{100-4\sqrt{445}}{8}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-100)+4\sqrt{445}}{2*4}=\frac{100+4\sqrt{445}}{8}
$