(7p+5)(p+2)=0

Solution for (7p+5)(p+2)=0 equation:

(7p+5)(p+2)=0

We multiply parentheses ..

(+7p^2+14p+5p+10)=0

We get rid of parentheses

7p^2+14p+5p+10=0

We add all the numbers together, and all the variables

7p^2+19p+10=0

a = 7; b = 19; c = +10;
Δ = b2-4ac
Δ = 192-4·7·10
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-9}{2*7}=\frac{-28}{14}
=-2
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+9}{2*7}=\frac{-10}{14}
=-5/7
$