5=9z+2z^2

Solution for 5=9z+2z^2 equation:

5=9z+2z^2

We move all terms to the left:

5-(9z+2z^2)=0

We get rid of parentheses

-2z^2-9z+5=0

a = -2; b = -9; c = +5;
Δ = b2-4ac
Δ = -92-4·(-2)·5
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-11}{2*-2}=\frac{-2}{-4}
=1/2
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+11}{2*-2}=\frac{20}{-4}
=-5
$