3x^2+18x-123=0

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Solution for 3x^2+18x-123=0 equation:



3x^2+18x-123=0
a = 3; b = 18; c = -123;
Δ = b2-4ac
Δ = 182-4·3·(-123)
Δ = 1800
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{1800}=\sqrt{900*2}=\sqrt{900}*\sqrt{2}=30\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-30\sqrt{2}}{2*3}=\frac{-18-30\sqrt{2}}{6} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+30\sqrt{2}}{2*3}=\frac{-18+30\sqrt{2}}{6} $

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