(10x-4)(3x+8)=-0

Solution for (10x-4)(3x+8)=-0 equation:

(10x-4)(3x+8)=-0

We multiply parentheses ..

(+30x^2+80x-12x-32)=0

We get rid of parentheses

30x^2+80x-12x-32=0

We add all the numbers together, and all the variables

30x^2+68x-32=0

a = 30; b = 68; c = -32;
Δ = b2-4ac
Δ = 682-4·30·(-32)
Δ = 8464
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{8464}=92$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(68)-92}{2*30}=\frac{-160}{60}
=-2+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(68)+92}{2*30}=\frac{24}{60}
=2/5
$