4n(n-5)+2=3(n-1)

Solution for 4n(n-5)+2=3(n-1) equation:

4n(n-5)+2=3(n-1)

We move all terms to the left:

4n(n-5)+2-(3(n-1))=0

We multiply parentheses

4n^2-20n-(3(n-1))+2=0


We calculate terms in parentheses: -(3(n-1)), so:

3(n-1)

We multiply parentheses

3n-3

Back to the equation:

-(3n-3)


We get rid of parentheses

4n^2-20n-3n+3+2=0

We add all the numbers together, and all the variables

4n^2-23n+5=0

a = 4; b = -23; c = +5;
Δ = b2-4ac
Δ = -232-4·4·5
Δ = 449
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{449}}{2*4}=\frac{23-\sqrt{449}}{8}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+\sqrt{449}}{2*4}=\frac{23+\sqrt{449}}{8}
$