z=7z^2+43z-42

Solution for z=7z^2+43z-42 equation:

z=7z^2+43z-42

We move all terms to the left:

z-(7z^2+43z-42)=0

We get rid of parentheses

-7z^2+z-43z+42=0

We add all the numbers together, and all the variables

-7z^2-42z+42=0

a = -7; b = -42; c = +42;
Δ = b2-4ac
Δ = -422-4·(-7)·42
Δ = 2940
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:

$\sqrt{\Delta}=\sqrt{2940}=\sqrt{196*15}=\sqrt{196}*\sqrt{15}=14\sqrt{15}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-14\sqrt{15}}{2*-7}=\frac{42-14\sqrt{15}}{-14}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+14\sqrt{15}}{2*-7}=\frac{42+14\sqrt{15}}{-14}
$