-9z=-8z2+9

Solution for -9z=-8z2+9 equation:

-9z=-8z^2+9

We move all terms to the left:

-9z-(-8z^2+9)=0

We get rid of parentheses

8z^2-9z-9=0

a = 8; b = -9; c = -9;
Δ = b2-4ac
Δ = -92-4·8·(-9)
Δ = 369
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:

$\sqrt{\Delta}=\sqrt{369}=\sqrt{9*41}=\sqrt{9}*\sqrt{41}=3\sqrt{41}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-3\sqrt{41}}{2*8}=\frac{9-3\sqrt{41}}{16}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+3\sqrt{41}}{2*8}=\frac{9+3\sqrt{41}}{16}
$