-2z^2+15z+27=0

Solution for -2z^2+15z+27=0 equation:

-2z^2+15z+27=0

a = -2; b = 15; c = +27;
Δ = b2-4ac
Δ = 152-4·(-2)·27
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-21}{2*-2}=\frac{-36}{-4}
=+9
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+21}{2*-2}=\frac{6}{-4}
=-1+1/2
$