(4r+5)(2r-3)=0

Solution for (4r+5)(2r-3)=0 equation:

(4r+5)(2r-3)=0

We multiply parentheses ..

(+8r^2-12r+10r-15)=0

We get rid of parentheses

8r^2-12r+10r-15=0

We add all the numbers together, and all the variables

8r^2-2r-15=0

a = 8; b = -2; c = -15;
Δ = b2-4ac
Δ = -22-4·8·(-15)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-22}{2*8}=\frac{-20}{16}
=-1+1/4
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+22}{2*8}=\frac{24}{16}
=1+1/2
$