-9z(4-3z)=-(6-3z)-7

Solution for -9z(4-3z)=-(6-3z)-7 equation:

-9z(4-3z)=-(6-3z)-7

We move all terms to the left:

-9z(4-3z)-(-(6-3z)-7)=0

We add all the numbers together, and all the variables

-9z(-3z+4)-(-(-3z+6)-7)=0

We multiply parentheses

27z^2-36z-(-(-3z+6)-7)=0


We calculate terms in parentheses: -(-(-3z+6)-7), so:

-(-3z+6)-7

We get rid of parentheses

3z-6-7

We add all the numbers together, and all the variables

3z-13

Back to the equation:

-(3z-13)


We get rid of parentheses

27z^2-36z-3z+13=0

We add all the numbers together, and all the variables

27z^2-39z+13=0

a = 27; b = -39; c = +13;
Δ = b2-4ac
Δ = -392-4·27·13
Δ = 117
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:

$\sqrt{\Delta}=\sqrt{117}=\sqrt{9*13}=\sqrt{9}*\sqrt{13}=3\sqrt{13}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-39)-3\sqrt{13}}{2*27}=\frac{39-3\sqrt{13}}{54}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-39)+3\sqrt{13}}{2*27}=\frac{39+3\sqrt{13}}{54}
$