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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+2z^2+5z+14z+35)=0

We get rid of parentheses

2z^2+5z+14z+35=0

We add all the numbers together, and all the variables

2z^2+19z+35=0

a = 2; b = 19; c = +35;
Δ = b2-4ac
Δ = 192-4·2·35
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-9}{2*2}=\frac{-28}{4}
=-7
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+9}{2*2}=\frac{-10}{4}
=-2+1/2
$