3x^2-212x+2304=0

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Solution for 3x^2-212x+2304=0 equation:



3x^2-212x+2304=0
a = 3; b = -212; c = +2304;
Δ = b2-4ac
Δ = -2122-4·3·2304
Δ = 17296
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:
$\sqrt{\Delta}=\sqrt{17296}=\sqrt{16*1081}=\sqrt{16}*\sqrt{1081}=4\sqrt{1081}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-212)-4\sqrt{1081}}{2*3}=\frac{212-4\sqrt{1081}}{6} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-212)+4\sqrt{1081}}{2*3}=\frac{212+4\sqrt{1081}}{6} $

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