4z+56z+52z+1=z2+z

Solution for 4z+56z+52z+1=z2+z equation:

4z+56z+52z+1=z2+z

We move all terms to the left:

4z+56z+52z+1-(z2+z)=0

We add all the numbers together, and all the variables

-(+z^2+z)+4z+56z+52z+1=0

We add all the numbers together, and all the variables

-(+z^2+z)+112z+1=0

We get rid of parentheses

-z^2-z+112z+1=0

We add all the numbers together, and all the variables

-1z^2+111z+1=0

a = -1; b = 111; c = +1;
Δ = b2-4ac
Δ = 1112-4·(-1)·1
Δ = 12325
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{12325}=\sqrt{25*493}=\sqrt{25}*\sqrt{493}=5\sqrt{493}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(111)-5\sqrt{493}}{2*-1}=\frac{-111-5\sqrt{493}}{-2}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(111)+5\sqrt{493}}{2*-1}=\frac{-111+5\sqrt{493}}{-2}
$