(x+5)(x-5)+1=(x+6)

Solution for (x+5)(x-5)+1=(x+6) equation:

(x+5)(x-5)+1=(x+6)

We move all terms to the left:

(x+5)(x-5)+1-((x+6))=0

We use the square of the difference formula

x^2-((x+6))-25+1=0


We calculate terms in parentheses: -((x+6)), so:

(x+6)

We get rid of parentheses

x+6

Back to the equation:

-(x+6)


We add all the numbers together, and all the variables

x^2-(x+6)-24=0

We get rid of parentheses

x^2-x-6-24=0

We add all the numbers together, and all the variables

x^2-1x-30=0

a = 1; b = -1; c = -30;
Δ = b2-4ac
Δ = -12-4·1·(-30)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-11}{2*1}=\frac{-10}{2}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+11}{2*1}=\frac{12}{2}
=6
$