n^2+5n-1=3n+2(n+25)

Solution for n^2+5n-1=3n+2(n+25) equation:

n^2+5n-1=3n+2(n+25)

We move all terms to the left:

n^2+5n-1-(3n+2(n+25))=0


We calculate terms in parentheses: -(3n+2(n+25)), so:

3n+2(n+25)

We multiply parentheses

3n+2n+50

We add all the numbers together, and all the variables

5n+50

Back to the equation:

-(5n+50)


We get rid of parentheses

n^2+5n-5n-50-1=0

We add all the numbers together, and all the variables

n^2-51=0

a = 1; b = 0; c = -51;
Δ = b2-4ac
Δ = 02-4·1·(-51)
Δ = 204
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:

$\sqrt{\Delta}=\sqrt{204}=\sqrt{4*51}=\sqrt{4}*\sqrt{51}=2\sqrt{51}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{51}}{2*1}=\frac{0-2\sqrt{51}}{2}
=-\frac{2\sqrt{51}}{2}
=-\sqrt{51}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{51}}{2*1}=\frac{0+2\sqrt{51}}{2}
=\frac{2\sqrt{51}}{2}
=\sqrt{51}
$