3x^2+33x-135=0

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Solution for 3x^2+33x-135=0 equation:



3x^2+33x-135=0
a = 3; b = 33; c = -135;
Δ = b2-4ac
Δ = 332-4·3·(-135)
Δ = 2709
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:
$\sqrt{\Delta}=\sqrt{2709}=\sqrt{9*301}=\sqrt{9}*\sqrt{301}=3\sqrt{301}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-3\sqrt{301}}{2*3}=\frac{-33-3\sqrt{301}}{6} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+3\sqrt{301}}{2*3}=\frac{-33+3\sqrt{301}}{6} $

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