3x^2+17x+18=0

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Solution for 3x^2+17x+18=0 equation: 



3x^2+17x+18=0

a = 3; b = 17; c = +18;
Δ = b2-4ac
Δ = 172-4·3·18
Δ = 73
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{-(17)-\sqrt{73}}{2*3}=\frac{-17-\sqrt{73}}{6}
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+\sqrt{73}}{2*3}=\frac{-17+\sqrt{73}}{6}
$

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