1400=10z2+2

Solution for 1400=10z2+2 equation:

1400=10z^2+2

We move all terms to the left:

1400-(10z^2+2)=0

We get rid of parentheses

-10z^2-2+1400=0

We add all the numbers together, and all the variables

-10z^2+1398=0

a = -10; b = 0; c = +1398;
Δ = b2-4ac
Δ = 02-4·(-10)·1398
Δ = 55920
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:

$\sqrt{\Delta}=\sqrt{55920}=\sqrt{16*3495}=\sqrt{16}*\sqrt{3495}=4\sqrt{3495}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{3495}}{2*-10}=\frac{0-4\sqrt{3495}}{-20}
=-\frac{4\sqrt{3495}}{-20}
=-\frac{\sqrt{3495}}{-5}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{3495}}{2*-10}=\frac{0+4\sqrt{3495}}{-20}
=\frac{4\sqrt{3495}}{-20}
=\frac{\sqrt{3495}}{-5}
$