3x^2-53x+232=0

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Solution for 3x^2-53x+232=0 equation:



3x^2-53x+232=0
a = 3; b = -53; c = +232;
Δ = b2-4ac
Δ = -532-4·3·232
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-53)-5}{2*3}=\frac{48}{6} =8 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-53)+5}{2*3}=\frac{58}{6} =9+2/3 $

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