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

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

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