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

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

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

We multiply parentheses ..

(+8r^2+10r-20r-25)=0

We get rid of parentheses

8r^2+10r-20r-25=0

We add all the numbers together, and all the variables

8r^2-10r-25=0

a = 8; b = -10; c = -25;
Δ = b2-4ac
Δ = -102-4·8·(-25)
Δ = 900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{900}=30$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-30}{2*8}=\frac{-20}{16}
=-1+1/4
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+30}{2*8}=\frac{40}{16}
=2+1/2
$