18z^2-31z+12=-2z^2

Solution for 18z^2-31z+12=-2z^2 equation:

18z^2-31z+12=-2z^2

We move all terms to the left:

18z^2-31z+12-(-2z^2)=0

We get rid of parentheses

18z^2+2z^2-31z+12=0

We add all the numbers together, and all the variables

20z^2-31z+12=0

a = 20; b = -31; c = +12;
Δ = b2-4ac
Δ = -312-4·20·12
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-31)-1}{2*20}=\frac{30}{40}
=3/4
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+1}{2*20}=\frac{32}{40}
=4/5
$