48z^2+52z+12=0

Solution for 48z^2+52z+12=0 equation:

48z^2+52z+12=0

a = 48; b = 52; c = +12;
Δ = b2-4ac
Δ = 522-4·48·12
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(52)-20}{2*48}=\frac{-72}{96}
=-3/4
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(52)+20}{2*48}=\frac{-32}{96}
=-1/3
$