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

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

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

We multiply parentheses ..

(+49z^2+35z+7z+5)=0

We get rid of parentheses

49z^2+35z+7z+5=0

We add all the numbers together, and all the variables

49z^2+42z+5=0

a = 49; b = 42; c = +5;
Δ = b2-4ac
Δ = 422-4·49·5
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-28}{2*49}=\frac{-70}{98}
=-5/7
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+28}{2*49}=\frac{-14}{98}
=-1/7
$