17z=6z^2+10

Solution for 17z=6z^2+10 equation:

17z=6z^2+10

We move all terms to the left:

17z-(6z^2+10)=0

We get rid of parentheses

-6z^2+17z-10=0

a = -6; b = 17; c = -10;
Δ = b2-4ac
Δ = 172-4·(-6)·(-10)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-7}{2*-6}=\frac{-24}{-12}
=+2
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+7}{2*-6}=\frac{-10}{-12}
=5/6
$