(4r+2)(2r-3)=0

Solution for (4r+2)(2r-3)=0 equation:

(4r+2)(2r-3)=0

We multiply parentheses ..

(+8r^2-12r+4r-6)=0

We get rid of parentheses

8r^2-12r+4r-6=0

We add all the numbers together, and all the variables

8r^2-8r-6=0

a = 8; b = -8; c = -6;
Δ = b2-4ac
Δ = -82-4·8·(-6)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-16}{2*8}=\frac{-8}{16}
=-1/2
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+16}{2*8}=\frac{24}{16}
=1+1/2
$