2(20x^2+x-12)=0

Solution for 2(20x^2+x-12)=0 equation:

2(20x^2+x-12)=0

We multiply parentheses

40x^2+2x-24=0

a = 40; b = 2; c = -24;
Δ = b2-4ac
Δ = 22-4·40·(-24)
Δ = 3844
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3844}=62$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-62}{2*40}=\frac{-64}{80}
=-4/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+62}{2*40}=\frac{60}{80}
=3/4
$