(5z-3)(7+z)=0

Solution for (5z-3)(7+z)=0 equation:

(5z-3)(7+z)=0

We add all the numbers together, and all the variables

(5z-3)(z+7)=0

We multiply parentheses ..

(+5z^2+35z-3z-21)=0

We get rid of parentheses

5z^2+35z-3z-21=0

We add all the numbers together, and all the variables

5z^2+32z-21=0

a = 5; b = 32; c = -21;
Δ = b2-4ac
Δ = 322-4·5·(-21)
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1444}=38$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-38}{2*5}=\frac{-70}{10}
=-7
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+38}{2*5}=\frac{6}{10}
=3/5
$