1=4z+8z2

Solution for 1=4z+8z2 equation:

1=4z+8z^2

We move all terms to the left:

1-(4z+8z^2)=0

We get rid of parentheses

-8z^2-4z+1=0

a = -8; b = -4; c = +1;
Δ = b2-4ac
Δ = -42-4·(-8)·1
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{3}}{2*-8}=\frac{4-4\sqrt{3}}{-16}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{3}}{2*-8}=\frac{4+4\sqrt{3}}{-16}
$