(3x-12)(x=4)=16

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Solution for (3x-12)(x=4)=16 equation: 



(3x-12)(x=4)=16

We move all terms to the left:

(3x-12)(x-(4))=0

We multiply parentheses ..

(+3x^2-12x-12x+48)=0

We get rid of parentheses

3x^2-12x-12x+48=0

We add all the numbers together, and all the variables

3x^2-24x+48=0

a = 3; b = -24; c = +48;
Δ = b2-4ac
Δ = -242-4·3·48
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{24}{6}=4$

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