2z^2+z+3=4

Solution for 2z^2+z+3=4 equation:

2z^2+z+3=4

We move all terms to the left:

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

We add all the numbers together, and all the variables

2z^2+z-1=0

a = 2; b = 1; c = -1;
Δ = b2-4ac
Δ = 12-4·2·(-1)
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-3}{2*2}=\frac{-4}{4}
=-1
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+3}{2*2}=\frac{2}{4}
=1/2
$