2x^2+19x+17=0

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Solution for 2x^2+19x+17=0 equation:



2x^2+19x+17=0
a = 2; b = 19; c = +17;
Δ = b2-4ac
Δ = 192-4·2·17
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{-(19)-15}{2*2}=\frac{-34}{4} =-8+1/2 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+15}{2*2}=\frac{-4}{4} =-1 $

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