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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2z^2+9z-35=0

a = 2; b = 9; c = -35;
Δ = b2-4ac
Δ = 92-4·2·(-35)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-19}{2*2}=\frac{-28}{4}
=-7
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+19}{2*2}=\frac{10}{4}
=2+1/2
$