28z^2+47z=0

Solution for 28z^2+47z=0 equation:

28z^2+47z=0

a = 28; b = 47; c = 0;
Δ = b2-4ac
Δ = 472-4·28·0
Δ = 2209
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2209}=47$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(47)-47}{2*28}=\frac{-94}{56}
=-1+19/28
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(47)+47}{2*28}=\frac{0}{56}
=0
$