2x^2+20x-528=0

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Solution for 2x^2+20x-528=0 equation:



2x^2+20x-528=0
a = 2; b = 20; c = -528;
Δ = b2-4ac
Δ = 202-4·2·(-528)
Δ = 4624
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4624}=68$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-68}{2*2}=\frac{-88}{4} =-22 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+68}{2*2}=\frac{48}{4} =12 $

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