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

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

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

We multiply parentheses ..

(+3r^2+21r-5r-35)=0

We get rid of parentheses

3r^2+21r-5r-35=0

We add all the numbers together, and all the variables

3r^2+16r-35=0

a = 3; b = 16; c = -35;
Δ = b2-4ac
Δ = 162-4·3·(-35)
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-26}{2*3}=\frac{-42}{6}
=-7
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+26}{2*3}=\frac{10}{6}
=1+2/3
$