2^4=6z2+7

Solution for 2^4=6z2+7 equation:

2^4=6z^2+7

We move all terms to the left:

2^4-(6z^2+7)=0

We add all the numbers together, and all the variables

-(6z^2+7)+16=0

We get rid of parentheses

-6z^2-7+16=0

We add all the numbers together, and all the variables

-6z^2+9=0

a = -6; b = 0; c = +9;
Δ = b2-4ac
Δ = 02-4·(-6)·9
Δ = 216
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:

$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{6}}{2*-6}=\frac{0-6\sqrt{6}}{-12}
=-\frac{6\sqrt{6}}{-12}
=-\frac{\sqrt{6}}{-2}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{6}}{2*-6}=\frac{0+6\sqrt{6}}{-12}
=\frac{6\sqrt{6}}{-12}
=\frac{\sqrt{6}}{-2}
$