(3x-8)(2x+5)=0

Solution for (3x-8)(2x+5)=0 equation:

(3x-8)(2x+5)=0

We multiply parentheses ..

(+6x^2+15x-16x-40)=0

We get rid of parentheses

6x^2+15x-16x-40=0

We add all the numbers together, and all the variables

6x^2-1x-40=0

a = 6; b = -1; c = -40;
Δ = b2-4ac
Δ = -12-4·6·(-40)
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-31}{2*6}=\frac{-30}{12}
=-2+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+31}{2*6}=\frac{32}{12}
=2+2/3
$