4z(5z+3)=114

Solution for 4z(5z+3)=114 equation:

4z(5z+3)=114

We move all terms to the left:

4z(5z+3)-(114)=0

We multiply parentheses

20z^2+12z-114=0

a = 20; b = 12; c = -114;
Δ = b2-4ac
Δ = 122-4·20·(-114)
Δ = 9264
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{9264}=\sqrt{16*579}=\sqrt{16}*\sqrt{579}=4\sqrt{579}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{579}}{2*20}=\frac{-12-4\sqrt{579}}{40}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{579}}{2*20}=\frac{-12+4\sqrt{579}}{40}
$