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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-4x^2+6x-10x+15)=0

We get rid of parentheses

-4x^2+6x-10x+15=0

We add all the numbers together, and all the variables

-4x^2-4x+15=0

a = -4; b = -4; c = +15;
Δ = b2-4ac
Δ = -42-4·(-4)·15
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-16}{2*-4}=\frac{-12}{-8}
=1+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+16}{2*-4}=\frac{20}{-8}
=-2+1/2
$