35(5+10n)=5(n+2)n=

Solution for 35(5+10n)=5(n+2)n= equation:

35(5+10n)=5(n+2)n=

We move all terms to the left:

35(5+10n)-(5(n+2)n)=0

We add all the numbers together, and all the variables

35(10n+5)-(5(n+2)n)=0

We multiply parentheses

350n-(5(n+2)n)+175=0


We calculate terms in parentheses: -(5(n+2)n), so:

5(n+2)n

We multiply parentheses

5n^2+10n

Back to the equation:

-(5n^2+10n)


We get rid of parentheses

-5n^2+350n-10n+175=0

We add all the numbers together, and all the variables

-5n^2+340n+175=0

a = -5; b = 340; c = +175;
Δ = b2-4ac
Δ = 3402-4·(-5)·175
Δ = 119100
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:

$\sqrt{\Delta}=\sqrt{119100}=\sqrt{100*1191}=\sqrt{100}*\sqrt{1191}=10\sqrt{1191}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(340)-10\sqrt{1191}}{2*-5}=\frac{-340-10\sqrt{1191}}{-10}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(340)+10\sqrt{1191}}{2*-5}=\frac{-340+10\sqrt{1191}}{-10}
$