2w+3=4w^2+14w+12

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Solution for 2w+3=4w^2+14w+12 equation: 



2w+3=4w^2+14w+12

We move all terms to the left:

2w+3-(4w^2+14w+12)=0

We get rid of parentheses

-4w^2+2w-14w-12+3=0

We add all the numbers together, and all the variables

-4w^2-12w-9=0

a = -4; b = -12; c = -9;
Δ = b2-4ac
Δ = -122-4·(-4)·(-9)
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$w=\frac{-b}{2a}=\frac{12}{-8}=-1+1/2$

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