20x+2(3-8x^2)=0

Solution for 20x+2(3-8x^2)=0 equation:

20x+2(3-8x^2)=0

We multiply parentheses

-16x^2+20x+6=0

a = -16; b = 20; c = +6;
Δ = b2-4ac
Δ = 202-4·(-16)·6
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-28}{2*-16}=\frac{-48}{-32}
=1+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+28}{2*-16}=\frac{8}{-32}
=-1/4
$