9n+48=7n(n-2)

Solution for 9n+48=7n(n-2) equation:

9n+48=7n(n-2)

We move all terms to the left:

9n+48-(7n(n-2))=0


We calculate terms in parentheses: -(7n(n-2)), so:

7n(n-2)

We multiply parentheses

7n^2-14n

Back to the equation:

-(7n^2-14n)


We get rid of parentheses

-7n^2+9n+14n+48=0

We add all the numbers together, and all the variables

-7n^2+23n+48=0

a = -7; b = 23; c = +48;
Δ = b2-4ac
Δ = 232-4·(-7)·48
Δ = 1873
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-\sqrt{1873}}{2*-7}=\frac{-23-\sqrt{1873}}{-14}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+\sqrt{1873}}{2*-7}=\frac{-23+\sqrt{1873}}{-14}
$