(2z+4)*(-2z^2+4z+3=)

Solution for (2z+4)*(-2z^2+4z+3=) equation:

(2z+4)(-2z^2+4z+3=)

We move all terms to the left:

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


We calculate terms in parentheses: +(2z+4)(-2z^2+4z+3-()), so:

2z+4)(-2z^2+4z+3-()

determiningTheFunctionDomain
-2z^2+2z+4z+3+4)(-()

We add all the numbers together, and all the variables

-2z^2+6z

Back to the equation:

+(-2z^2+6z)


We get rid of parentheses

-2z^2+6z=0

a = -2; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·(-2)·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*-2}=\frac{-12}{-4}
=+3
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*-2}=\frac{0}{-4}
=0
$