10(z+1)=5(z-3)z=

Solution for 10(z+1)=5(z-3)z= equation:

10(z+1)=5(z-3)z=

We move all terms to the left:

10(z+1)-(5(z-3)z)=0

We multiply parentheses

10z-(5(z-3)z)+10=0


We calculate terms in parentheses: -(5(z-3)z), so:

5(z-3)z

We multiply parentheses

5z^2-15z

Back to the equation:

-(5z^2-15z)


We get rid of parentheses

-5z^2+10z+15z+10=0

We add all the numbers together, and all the variables

-5z^2+25z+10=0

a = -5; b = 25; c = +10;
Δ = b2-4ac
Δ = 252-4·(-5)·10
Δ = 825
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:

$\sqrt{\Delta}=\sqrt{825}=\sqrt{25*33}=\sqrt{25}*\sqrt{33}=5\sqrt{33}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-5\sqrt{33}}{2*-5}=\frac{-25-5\sqrt{33}}{-10}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+5\sqrt{33}}{2*-5}=\frac{-25+5\sqrt{33}}{-10}
$