7=z(3z-4)

Solution for 7=z(3z-4) equation:

7=z(3z-4)

We move all terms to the left:

7-(z(3z-4))=0


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

z(3z-4)

We multiply parentheses

3z^2-4z

Back to the equation:

-(3z^2-4z)


We get rid of parentheses

-3z^2+4z+7=0

a = -3; b = 4; c = +7;
Δ = b2-4ac
Δ = 42-4·(-3)·7
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-10}{2*-3}=\frac{-14}{-6}
=2+1/3
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+10}{2*-3}=\frac{6}{-6}
=-1
$