(8r+7)(r+2)=0

Solution for (8r+7)(r+2)=0 equation:

(8r+7)(r+2)=0

We multiply parentheses ..

(+8r^2+16r+7r+14)=0

We get rid of parentheses

8r^2+16r+7r+14=0

We add all the numbers together, and all the variables

8r^2+23r+14=0

a = 8; b = 23; c = +14;
Δ = b2-4ac
Δ = 232-4·8·14
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-9}{2*8}=\frac{-32}{16}
=-2
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+9}{2*8}=\frac{-14}{16}
=-7/8
$