9z+(z+2)=10+z2

Solution for 9z+(z+2)=10+z2 equation:

9z+(z+2)=10+z2

We move all terms to the left:

9z+(z+2)-(10+z2)=0

We add all the numbers together, and all the variables

-(+z^2+10)+9z+(z+2)=0

We get rid of parentheses

-z^2+9z+z-10+2=0

We add all the numbers together, and all the variables

-1z^2+10z-8=0

a = -1; b = 10; c = -8;
Δ = b2-4ac
Δ = 102-4·(-1)·(-8)
Δ = 68
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:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{17}}{2*-1}=\frac{-10-2\sqrt{17}}{-2}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{17}}{2*-1}=\frac{-10+2\sqrt{17}}{-2}
$