4(z+3)(z-1)1=36

Solution for 4(z+3)(z-1)1=36 equation:

4(z+3)(z-1)1=36

We move all terms to the left:

4(z+3)(z-1)1-(36)=0

We multiply parentheses ..

4(+z^2-1z+3z-3)1-36=0

We multiply parentheses

4z^2-4z+12z-12-36=0

We add all the numbers together, and all the variables

4z^2+8z-48=0

a = 4; b = 8; c = -48;
Δ = b2-4ac
Δ = 82-4·4·(-48)
Δ = 832
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{832}=\sqrt{64*13}=\sqrt{64}*\sqrt{13}=8\sqrt{13}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{13}}{2*4}=\frac{-8-8\sqrt{13}}{8}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{13}}{2*4}=\frac{-8+8\sqrt{13}}{8}
$