7z^2-50=13

Solution for 7z^2-50=13 equation:

7z^2-50=13

We move all terms to the left:

7z^2-50-(13)=0

We add all the numbers together, and all the variables

7z^2-63=0

a = 7; b = 0; c = -63;
Δ = b2-4ac
Δ = 02-4·7·(-63)
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-42}{2*7}=\frac{-42}{14}
=-3
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+42}{2*7}=\frac{42}{14}
=3
$