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

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Solution for (x-5)(x+3)+16=0 equation: 



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

We multiply parentheses ..

(+x^2+3x-5x-15)+16=0

We get rid of parentheses

x^2+3x-5x-15+16=0

We add all the numbers together, and all the variables

x^2-2x+1=0

a = 1; b = -2; c = +1;
Δ = b2-4ac
Δ = -22-4·1·1
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{2}{2}=1$

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