z^2+9z+81/4=1+81/4

Solution for z^2+9z+81/4=1+81/4 equation:

z^2+9z+81/4=1+81/4

We move all terms to the left:

z^2+9z+81/4-(1+81/4)=0

We add all the numbers together, and all the variables

z^2+9z+81/4-(81/4+1)=0

We get rid of parentheses

z^2+9z-1+81/4-81/4=0

We multiply all the terms by the denominator


z^2*4+9z*4+81-81-1*4=0

We add all the numbers together, and all the variables

z^2*4+9z*4-4=0

Wy multiply elements

4z^2+36z-4=0

a = 4; b = 36; c = -4;
Δ = b2-4ac
Δ = 362-4·4·(-4)
Δ = 1360
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:

$\sqrt{\Delta}=\sqrt{1360}=\sqrt{16*85}=\sqrt{16}*\sqrt{85}=4\sqrt{85}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-4\sqrt{85}}{2*4}=\frac{-36-4\sqrt{85}}{8}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+4\sqrt{85}}{2*4}=\frac{-36+4\sqrt{85}}{8}
$