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

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

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

We move all terms to the left:

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

We get rid of parentheses

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


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

(n^2-25)

We get rid of parentheses

n^2-25

Back to the equation:

-(n^2-25)


We get rid of parentheses

-n^2-3n+25+3=0

We add all the numbers together, and all the variables

-1n^2-3n+28=0

a = -1; b = -3; c = +28;
Δ = b2-4ac
Δ = -32-4·(-1)·28
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-11}{2*-1}=\frac{-8}{-2}
=+4
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+11}{2*-1}=\frac{14}{-2}
=-7
$