4b^2+19b+17=0

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Solution for 4b^2+19b+17=0 equation: 



4b^2+19b+17=0

a = 4; b = 19; c = +17;
Δ = b2-4ac
Δ = 192-4·4·17
Δ = 89
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-\sqrt{89}}{2*4}=\frac{-19-\sqrt{89}}{8}
$


$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{89}}{2*4}=\frac{-19+\sqrt{89}}{8}
$

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