(x+5)(x-5)+16=(x-3)

Solution for (x+5)(x-5)+16=(x-3) equation:

(x+5)(x-5)+16=(x-3)

We move all terms to the left:

(x+5)(x-5)+16-((x-3))=0

We use the square of the difference formula

x^2-((x-3))-25+16=0


We calculate terms in parentheses: -((x-3)), so:

(x-3)

We get rid of parentheses

x-3

Back to the equation:

-(x-3)


We add all the numbers together, and all the variables

x^2-(x-3)-9=0

We get rid of parentheses

x^2-x+3-9=0

We add all the numbers together, and all the variables

x^2-1x-6=0

a = 1; b = -1; c = -6;
Δ = b2-4ac
Δ = -12-4·1·(-6)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-5}{2*1}=\frac{-4}{2}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+5}{2*1}=\frac{6}{2}
=3
$