2(3x^2+x-4)=0

Solution for 2(3x^2+x-4)=0 equation:

2(3x^2+x-4)=0

We multiply parentheses

6x^2+2x-8=0

a = 6; b = 2; c = -8;
Δ = b2-4ac
Δ = 22-4·6·(-8)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-14}{2*6}=\frac{-16}{12}
=-1+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+14}{2*6}=\frac{12}{12}
=1
$