4x^2+23x+19=0

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Solution for 4x^2+23x+19=0 equation:



4x^2+23x+19=0
a = 4; b = 23; c = +19;
Δ = b2-4ac
Δ = 232-4·4·19
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-15}{2*4}=\frac{-38}{8} =-4+3/4 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+15}{2*4}=\frac{-8}{8} =-1 $

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