2w^2+7w-3=0

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Solution for 2w^2+7w-3=0 equation:



2w^2+7w-3=0
a = 2; b = 7; c = -3;
Δ = b2-4ac
Δ = 72-4·2·(-3)
Δ = 73
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-\sqrt{73}}{2*2}=\frac{-7-\sqrt{73}}{4} $
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{73}}{2*2}=\frac{-7+\sqrt{73}}{4} $

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