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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

5z^2+33z-14=0

a = 5; b = 33; c = -14;
Δ = b2-4ac
Δ = 332-4·5·(-14)
Δ = 1369
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1369}=37$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-37}{2*5}=\frac{-70}{10}
=-7
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+37}{2*5}=\frac{4}{10}
=2/5
$