4z^2+z=1

Solution for 4z^2+z=1 equation:

4z^2+z=1

We move all terms to the left:

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

a = 4; b = 1; c = -1;
Δ = b2-4ac
Δ = 12-4·4·(-1)
Δ = 17
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{17}}{2*4}=\frac{-1-\sqrt{17}}{8}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{17}}{2*4}=\frac{-1+\sqrt{17}}{8}
$