-3x^2+x+102=0

Solution for -3x^2+x+102=0 equation:

-3x^2+x+102=0

a = -3; b = 1; c = +102;
Δ = b2-4ac
Δ = 12-4·(-3)·102
Δ = 1225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1225}=35$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-35}{2*-3}=\frac{-36}{-6}
=+6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+35}{2*-3}=\frac{34}{-6}
=-5+2/3
$