20n(n+2)=7n-10

Solution for 20n(n+2)=7n-10 equation:

20n(n+2)=7n-10

We move all terms to the left:

20n(n+2)-(7n-10)=0

We multiply parentheses

20n^2+40n-(7n-10)=0

We get rid of parentheses

20n^2+40n-7n+10=0

We add all the numbers together, and all the variables

20n^2+33n+10=0

a = 20; b = 33; c = +10;
Δ = b2-4ac
Δ = 332-4·20·10
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-17}{2*20}=\frac{-50}{40}
=-1+1/4
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+17}{2*20}=\frac{-16}{40}
=-2/5
$