(r+2)(8r-3)=0

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

(r+2)(8r-3)=0

We multiply parentheses ..

(+8r^2-3r+16r-6)=0

We get rid of parentheses

8r^2-3r+16r-6=0

We add all the numbers together, and all the variables

8r^2+13r-6=0

a = 8; b = 13; c = -6;
Δ = b2-4ac
Δ = 132-4·8·(-6)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-19}{2*8}=\frac{-32}{16}
=-2
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+19}{2*8}=\frac{6}{16}
=3/8
$