19x^2+38x=0

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Solution for 19x^2+38x=0 equation: 



19x^2+38x=0

a = 19; b = 38; c = 0;
Δ = b2-4ac
Δ = 382-4·19·0
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1444}=38$


$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(38)-38}{2*19}=\frac{-76}{38}
=-2
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(38)+38}{2*19}=\frac{0}{38}
=0
$

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