-3x^2+x+12=0

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Solution for -3x^2+x+12=0 equation:



-3x^2+x+12=0
a = -3; b = 1; c = +12;
Δ = b2-4ac
Δ = 12-4·(-3)·12
Δ = 145
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{-(1)-\sqrt{145}}{2*-3}=\frac{-1-\sqrt{145}}{-6} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{145}}{2*-3}=\frac{-1+\sqrt{145}}{-6} $

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