-8a^2+12a-4=0

Solution for -8a^2+12a-4=0 equation:

-8a^2+12a-4=0

a = -8; b = 12; c = -4;
Δ = b2-4ac
Δ = 122-4·(-8)·(-4)
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*-8}=\frac{-16}{-16}
=1
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4}{2*-8}=\frac{-8}{-16}
=1/2
$