4z+82=7z^2

Solution for 4z+82=7z^2 equation:

4z+82=7z^2

We move all terms to the left:

4z+82-(7z^2)=0

determiningTheFunctionDomain
-7z^2+4z+82=0

a = -7; b = 4; c = +82;
Δ = b2-4ac
Δ = 42-4·(-7)·82
Δ = 2312
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:

$\sqrt{\Delta}=\sqrt{2312}=\sqrt{1156*2}=\sqrt{1156}*\sqrt{2}=34\sqrt{2}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-34\sqrt{2}}{2*-7}=\frac{-4-34\sqrt{2}}{-14}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+34\sqrt{2}}{2*-7}=\frac{-4+34\sqrt{2}}{-14}
$