3x^2+19x+17=0

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Solution for 3x^2+19x+17=0 equation: 



3x^2+19x+17=0

a = 3; b = 19; c = +17;
Δ = b2-4ac
Δ = 192-4·3·17
Δ = 157
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-\sqrt{157}}{2*3}=\frac{-19-\sqrt{157}}{6}
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{157}}{2*3}=\frac{-19+\sqrt{157}}{6}
$

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