(8+z)(4z-7)=0

Solution for (8+z)(4z-7)=0 equation:

(8+z)(4z-7)=0

We add all the numbers together, and all the variables

(z+8)(4z-7)=0

We multiply parentheses ..

(+4z^2-7z+32z-56)=0

We get rid of parentheses

4z^2-7z+32z-56=0

We add all the numbers together, and all the variables

4z^2+25z-56=0

a = 4; b = 25; c = -56;
Δ = b2-4ac
Δ = 252-4·4·(-56)
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1521}=39$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-39}{2*4}=\frac{-64}{8}
=-8
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+39}{2*4}=\frac{14}{8}
=1+3/4
$