3x^2+x-30100=0

Solution for 3x^2+x-30100=0 equation:

3x^2+x-30100=0

a = 3; b = 1; c = -30100;
Δ = b2-4ac
Δ = 12-4·3·(-30100)
Δ = 361201
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361201}=601$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-601}{2*3}=\frac{-602}{6}
=-100+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+601}{2*3}=\frac{600}{6}
=100
$