(n+7)(11n-2)=(2n+1)(18n)

Solution for (n+7)(11n-2)=(2n+1)(18n) equation:

(n+7)(11n-2)=(2n+1)(18n)

We move all terms to the left:

(n+7)(11n-2)-((2n+1)(18n))=0

We multiply parentheses ..

(+11n^2-2n+77n-14)-((2n+1)18n)=0


We calculate terms in parentheses: -((2n+1)18n), so:

(2n+1)18n

We multiply parentheses

36n^2+18n

Back to the equation:

-(36n^2+18n)


We get rid of parentheses

11n^2-36n^2-2n+77n-18n-14=0

We add all the numbers together, and all the variables

-25n^2+57n-14=0

a = -25; b = 57; c = -14;
Δ = b2-4ac
Δ = 572-4·(-25)·(-14)
Δ = 1849
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1849}=43$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(57)-43}{2*-25}=\frac{-100}{-50}
=+2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(57)+43}{2*-25}=\frac{-14}{-50}
=7/25
$