3w2+22w+7=0

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Solution for 3w2+22w+7=0 equation:



3w^2+22w+7=0
a = 3; b = 22; c = +7;
Δ = b2-4ac
Δ = 222-4·3·7
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-20}{2*3}=\frac{-42}{6} =-7 $
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+20}{2*3}=\frac{-2}{6} =-1/3 $

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