4z2+11z+6=

Solution for 4z2+11z+6= equation:

4z^2+11z+6=

We move all terms to the left:

4z^2+11z+6-()=0

We add all the numbers together, and all the variables

4z^2+11z=0

a = 4; b = 11; c = 0;
Δ = b2-4ac
Δ = 112-4·4·0
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-11}{2*4}=\frac{-22}{8}
=-2+3/4
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+11}{2*4}=\frac{0}{8}
=0
$