(-3r-4)(5r+7)=0

Solution for (-3r-4)(5r+7)=0 equation:

(-3r-4)(5r+7)=0

We multiply parentheses ..

(-15r^2-21r-20r-28)=0

We get rid of parentheses

-15r^2-21r-20r-28=0

We add all the numbers together, and all the variables

-15r^2-41r-28=0

a = -15; b = -41; c = -28;
Δ = b2-4ac
Δ = -412-4·(-15)·(-28)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-41)-1}{2*-15}=\frac{40}{-30}
=-1+1/3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-41)+1}{2*-15}=\frac{42}{-30}
=-1+2/5
$