7z^2-38+15=0

Solution for 7z^2-38+15=0 equation:

7z^2-38+15=0

We add all the numbers together, and all the variables

7z^2-23=0

a = 7; b = 0; c = -23;
Δ = b2-4ac
Δ = 02-4·7·(-23)
Δ = 644
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:

$\sqrt{\Delta}=\sqrt{644}=\sqrt{4*161}=\sqrt{4}*\sqrt{161}=2\sqrt{161}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{161}}{2*7}=\frac{0-2\sqrt{161}}{14}
=-\frac{2\sqrt{161}}{14}
=-\frac{\sqrt{161}}{7}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{161}}{2*7}=\frac{0+2\sqrt{161}}{14}
=\frac{2\sqrt{161}}{14}
=\frac{\sqrt{161}}{7}
$