(4w^2)+(28w)=0

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Solution for (4w^2)+(28w)=0 equation: 



(4w^2)+(28w)=0

a = 4; b = 28; c = 0;
Δ = b2-4ac
Δ = 282-4·4·0
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$


$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-28}{2*4}=\frac{-56}{8}
=-7
$


$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+28}{2*4}=\frac{0}{8}
=0
$

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