5(x^2)+19x-4=0

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Solution for 5(x^2)+19x-4=0 equation: 



5(x^2)+19x-4=0

a = 5; b = 19; c = -4;
Δ = b2-4ac
Δ = 192-4·5·(-4)
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$


$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-21}{2*5}=\frac{-40}{10}
=-4
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+21}{2*5}=\frac{2}{10}
=1/5
$

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