4(3-5)+3(2)=z2-15

Solution for 4(3-5)+3(2)=z2-15 equation:

4(3-5)+3(2)=z2-15

We move all terms to the left:

4(3-5)+3(2)-(z2-15)=0

determiningTheFunctionDomain
-(z2-15)+32+4(3-5)=0

We add all the numbers together, and all the variables

-(+z^2-15)+32+4(-2)=0

We add all the numbers together, and all the variables

-(+z^2-15)+24=0

We get rid of parentheses

-z^2+15+24=0

We add all the numbers together, and all the variables

-1z^2+39=0

a = -1; b = 0; c = +39;
Δ = b2-4ac
Δ = 02-4·(-1)·39
Δ = 156
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:

$\sqrt{\Delta}=\sqrt{156}=\sqrt{4*39}=\sqrt{4}*\sqrt{39}=2\sqrt{39}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{39}}{2*-1}=\frac{0-2\sqrt{39}}{-2}
=-\frac{2\sqrt{39}}{-2}
=-\frac{\sqrt{39}}{-1}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{39}}{2*-1}=\frac{0+2\sqrt{39}}{-2}
=\frac{2\sqrt{39}}{-2}
=\frac{\sqrt{39}}{-1}
$