63z=49+15z2

Solution for 63z=49+15z2 equation:

63z=49+15z^2

We move all terms to the left:

63z-(49+15z^2)=0

We get rid of parentheses

-15z^2+63z-49=0

a = -15; b = 63; c = -49;
Δ = b2-4ac
Δ = 632-4·(-15)·(-49)
Δ = 1029
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:

$\sqrt{\Delta}=\sqrt{1029}=\sqrt{49*21}=\sqrt{49}*\sqrt{21}=7\sqrt{21}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(63)-7\sqrt{21}}{2*-15}=\frac{-63-7\sqrt{21}}{-30}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(63)+7\sqrt{21}}{2*-15}=\frac{-63+7\sqrt{21}}{-30}
$