(7+z)(5z+1)=0

Solution for (7+z)(5z+1)=0 equation:

(7+z)(5z+1)=0

We add all the numbers together, and all the variables

(z+7)(5z+1)=0

We multiply parentheses ..

(+5z^2+z+35z+7)=0

We get rid of parentheses

5z^2+z+35z+7=0

We add all the numbers together, and all the variables

5z^2+36z+7=0

a = 5; b = 36; c = +7;
Δ = b2-4ac
Δ = 362-4·5·7
Δ = 1156
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1156}=34$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-34}{2*5}=\frac{-70}{10}
=-7
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+34}{2*5}=\frac{-2}{10}
=-1/5
$