130=4z(3z+2)

Solution for 130=4z(3z+2) equation:

130=4z(3z+2)

We move all terms to the left:

130-(4z(3z+2))=0


We calculate terms in parentheses: -(4z(3z+2)), so:

4z(3z+2)

We multiply parentheses

12z^2+8z

Back to the equation:

-(12z^2+8z)


We get rid of parentheses

-12z^2-8z+130=0

a = -12; b = -8; c = +130;
Δ = b2-4ac
Δ = -82-4·(-12)·130
Δ = 6304
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:

$\sqrt{\Delta}=\sqrt{6304}=\sqrt{16*394}=\sqrt{16}*\sqrt{394}=4\sqrt{394}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4\sqrt{394}}{2*-12}=\frac{8-4\sqrt{394}}{-24}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4\sqrt{394}}{2*-12}=\frac{8+4\sqrt{394}}{-24}
$