100x^2+5.8x-0.928=0

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Solution for 100x^2+5.8x-0.928=0 equation:



100x^2+5.8x-0.928=0
a = 100; b = 5.8; c = -0.928;
Δ = b2-4ac
Δ = 5.82-4·100·(-0.928)
Δ = 404.84
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5.8)-\sqrt{404.84}}{2*100}=\frac{-5.8-\sqrt{404.84}}{200} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5.8)+\sqrt{404.84}}{2*100}=\frac{-5.8+\sqrt{404.84}}{200} $

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