(7z^2-31z+26)=0

Solution for (7z^2-31z+26)=0 equation:

(7z^2-31z+26)=0

We get rid of parentheses

7z^2-31z+26=0

a = 7; b = -31; c = +26;
Δ = b2-4ac
Δ = -312-4·7·26
Δ = 233
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-31)-\sqrt{233}}{2*7}=\frac{31-\sqrt{233}}{14}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+\sqrt{233}}{2*7}=\frac{31+\sqrt{233}}{14}
$