z^2+1=4+3z

Solution for z^2+1=4+3z equation:

z^2+1=4+3z

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

z^2-3z-4+1=0

We add all the numbers together, and all the variables

z^2-3z-3=0

a = 1; b = -3; c = -3;
Δ = b2-4ac
Δ = -32-4·1·(-3)
Δ = 21
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{21}}{2*1}=\frac{3-\sqrt{21}}{2}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{21}}{2*1}=\frac{3+\sqrt{21}}{2}
$