(7+z)(4z+3)=0

Solution for (7+z)(4z+3)=0 equation:

(7+z)(4z+3)=0

We add all the numbers together, and all the variables

(z+7)(4z+3)=0

We multiply parentheses ..

(+4z^2+3z+28z+21)=0

We get rid of parentheses

4z^2+3z+28z+21=0

We add all the numbers together, and all the variables

4z^2+31z+21=0

a = 4; b = 31; c = +21;
Δ = b2-4ac
Δ = 312-4·4·21
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(31)-25}{2*4}=\frac{-56}{8}
=-7
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+25}{2*4}=\frac{-6}{8}
=-3/4
$