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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

7z^2-32z+26=0

a = 7; b = -32; c = +26;
Δ = b2-4ac
Δ = -322-4·7·26
Δ = 296
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:

$\sqrt{\Delta}=\sqrt{296}=\sqrt{4*74}=\sqrt{4}*\sqrt{74}=2\sqrt{74}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-2\sqrt{74}}{2*7}=\frac{32-2\sqrt{74}}{14}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+2\sqrt{74}}{2*7}=\frac{32+2\sqrt{74}}{14}
$