2z^2=10z-6

Solution for 2z^2=10z-6 equation:

2z^2=10z-6

We move all terms to the left:

2z^2-(10z-6)=0

We get rid of parentheses

2z^2-10z+6=0

a = 2; b = -10; c = +6;
Δ = b2-4ac
Δ = -102-4·2·6
Δ = 52
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:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{13}}{2*2}=\frac{10-2\sqrt{13}}{4}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{13}}{2*2}=\frac{10+2\sqrt{13}}{4}
$