(8-3z)-(5z-3)+(5-2z)=(6z-4)(4z+2)

Solution for (8-3z)-(5z-3)+(5-2z)=(6z-4)(4z+2) equation:

(8-3z)-(5z-3)+(5-2z)=(6z-4)(4z+2)

We move all terms to the left:

(8-3z)-(5z-3)+(5-2z)-((6z-4)(4z+2))=0

We add all the numbers together, and all the variables

(-3z+8)-(5z-3)+(-2z+5)-((6z-4)(4z+2))=0

We get rid of parentheses

-3z-5z-2z-((6z-4)(4z+2))+8+3+5=0

We multiply parentheses ..

-((+24z^2+12z-16z-8))-3z-5z-2z+8+3+5=0


We calculate terms in parentheses: -((+24z^2+12z-16z-8)), so:

(+24z^2+12z-16z-8)

We get rid of parentheses

24z^2+12z-16z-8

We add all the numbers together, and all the variables

24z^2-4z-8

Back to the equation:

-(24z^2-4z-8)


We add all the numbers together, and all the variables

-10z-(24z^2-4z-8)+16=0

We get rid of parentheses

-24z^2-10z+4z+8+16=0

We add all the numbers together, and all the variables

-24z^2-6z+24=0

a = -24; b = -6; c = +24;
Δ = b2-4ac
Δ = -62-4·(-24)·24
Δ = 2340
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:

$\sqrt{\Delta}=\sqrt{2340}=\sqrt{36*65}=\sqrt{36}*\sqrt{65}=6\sqrt{65}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6\sqrt{65}}{2*-24}=\frac{6-6\sqrt{65}}{-48}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6\sqrt{65}}{2*-24}=\frac{6+6\sqrt{65}}{-48}
$