2x2+20x-4=0

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Solution for 2x2+20x-4=0 equation:



2x^2+20x-4=0
a = 2; b = 20; c = -4;
Δ = b2-4ac
Δ = 202-4·2·(-4)
Δ = 432
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{432}=\sqrt{144*3}=\sqrt{144}*\sqrt{3}=12\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-12\sqrt{3}}{2*2}=\frac{-20-12\sqrt{3}}{4} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+12\sqrt{3}}{2*2}=\frac{-20+12\sqrt{3}}{4} $

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