5z2-9=41+3z2

Solution for 5z2-9=41+3z2 equation:

5z^2-9=41+3z^2

We move all terms to the left:

5z^2-9-(41+3z^2)=0

We get rid of parentheses

5z^2-3z^2-41-9=0

We add all the numbers together, and all the variables

2z^2-50=0

a = 2; b = 0; c = -50;
Δ = b2-4ac
Δ = 02-4·2·(-50)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20}{2*2}=\frac{-20}{4}
=-5
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20}{2*2}=\frac{20}{4}
=5
$