(2r-5)(r+4)=0

Solution for (2r-5)(r+4)=0 equation:

(2r-5)(r+4)=0

We multiply parentheses ..

(+2r^2+8r-5r-20)=0

We get rid of parentheses

2r^2+8r-5r-20=0

We add all the numbers together, and all the variables

2r^2+3r-20=0

a = 2; b = 3; c = -20;
Δ = b2-4ac
Δ = 32-4·2·(-20)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-13}{2*2}=\frac{-16}{4}
=-4
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+13}{2*2}=\frac{10}{4}
=2+1/2
$