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3w(4w+1)=-16w-4
We move all terms to the left:
3w(4w+1)-(-16w-4)=0
We multiply parentheses
12w^2+3w-(-16w-4)=0
We get rid of parentheses
12w^2+3w+16w+4=0
We add all the numbers together, and all the variables
12w^2+19w+4=0
a = 12; b = 19; c = +4;
Δ = b2-4ac
Δ = 192-4·12·4
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{169}=13$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-13}{2*12}=\frac{-32}{24} =-1+1/3 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+13}{2*12}=\frac{-6}{24} =-1/4 $
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