2z^2+6=7z

Solution for 2z^2+6=7z equation:

2z^2+6=7z

We move all terms to the left:

2z^2+6-(7z)=0

a = 2; b = -7; c = +6;
Δ = b2-4ac
Δ = -72-4·2·6
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-1}{2*2}=\frac{6}{4}
=1+1/2
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+1}{2*2}=\frac{8}{4}
=2
$