-2a^2+12a-16=0

Solution for -2a^2+12a-16=0 equation:

-2a^2+12a-16=0

a = -2; b = 12; c = -16;
Δ = b2-4ac
Δ = 122-4·(-2)·(-16)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4}{2*-2}=\frac{-16}{-4}
=+4
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4}{2*-2}=\frac{-8}{-4}
=+2
$