15+28=3z^2-5

Solution for 15+28=3z^2-5 equation:

15+28=3z^2-5

We move all terms to the left:

15+28-(3z^2-5)=0

We add all the numbers together, and all the variables

-(3z^2-5)+43=0

We get rid of parentheses

-3z^2+5+43=0

We add all the numbers together, and all the variables

-3z^2+48=0

a = -3; b = 0; c = +48;
Δ = b2-4ac
Δ = 02-4·(-3)·48
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24}{2*-3}=\frac{-24}{-6}
=+4
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24}{2*-3}=\frac{24}{-6}
=-4
$