3z^2-12=3-(1+z)

Solution for 3z^2-12=3-(1+z) equation:

3z^2-12=3-(1+z)

We move all terms to the left:

3z^2-12-(3-(1+z))=0

We add all the numbers together, and all the variables

3z^2-(3-(z+1))-12=0


We calculate terms in parentheses: -(3-(z+1)), so:

3-(z+1)

determiningTheFunctionDomain
-(z+1)+3

We get rid of parentheses

-z-1+3

We add all the numbers together, and all the variables

-1z+2

Back to the equation:

-(-1z+2)


We get rid of parentheses

3z^2+1z-2-12=0

We add all the numbers together, and all the variables

3z^2+z-14=0

a = 3; b = 1; c = -14;
Δ = b2-4ac
Δ = 12-4·3·(-14)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-13}{2*3}=\frac{-14}{6}
=-2+1/3
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+13}{2*3}=\frac{12}{6}
=2
$