(5z-7)(9+z)=0

Solution for (5z-7)(9+z)=0 equation:

(5z-7)(9+z)=0

We add all the numbers together, and all the variables

(5z-7)(z+9)=0

We multiply parentheses ..

(+5z^2+45z-7z-63)=0

We get rid of parentheses

5z^2+45z-7z-63=0

We add all the numbers together, and all the variables

5z^2+38z-63=0

a = 5; b = 38; c = -63;
Δ = b2-4ac
Δ = 382-4·5·(-63)
Δ = 2704
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2704}=52$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(38)-52}{2*5}=\frac{-90}{10}
=-9
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(38)+52}{2*5}=\frac{14}{10}
=1+2/5
$