63z=49+18z2

Solution for 63z=49+18z2 equation:

63z=49+18z^2

We move all terms to the left:

63z-(49+18z^2)=0

We get rid of parentheses

-18z^2+63z-49=0

a = -18; b = 63; c = -49;
Δ = b2-4ac
Δ = 632-4·(-18)·(-49)
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(63)-21}{2*-18}=\frac{-84}{-36}
=2+1/3
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(63)+21}{2*-18}=\frac{-42}{-36}
=1+1/6
$