4z^2+62z=32

Solution for 4z^2+62z=32 equation:

4z^2+62z=32

We move all terms to the left:

4z^2+62z-(32)=0

a = 4; b = 62; c = -32;
Δ = b2-4ac
Δ = 622-4·4·(-32)
Δ = 4356
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4356}=66$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(62)-66}{2*4}=\frac{-128}{8}
=-16
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(62)+66}{2*4}=\frac{4}{8}
=1/2
$