3x^2+134x-4489=0

Solution for 3x^2+134x-4489=0 equation:

3x^2+134x-4489=0

a = 3; b = 134; c = -4489;
Δ = b2-4ac
Δ = 1342-4·3·(-4489)
Δ = 71824
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{71824}=268$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(134)-268}{2*3}=\frac{-402}{6}
=-67
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(134)+268}{2*3}=\frac{134}{6}
=22+1/3
$