20=2(z2+5)

Solution for 20=2(z2+5) equation:

20=2(z2+5)

We move all terms to the left:

20-(2(z2+5))=0

We add all the numbers together, and all the variables

-(2(+z^2+5))+20=0


We calculate terms in parentheses: -(2(+z^2+5)), so:

2(+z^2+5)

We multiply parentheses

2z^2+10

Back to the equation:

-(2z^2+10)


We get rid of parentheses

-2z^2-10+20=0

We add all the numbers together, and all the variables

-2z^2+10=0

a = -2; b = 0; c = +10;
Δ = b2-4ac
Δ = 02-4·(-2)·10
Δ = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{5}}{2*-2}=\frac{0-4\sqrt{5}}{-4}
=-\frac{4\sqrt{5}}{-4}
=-\frac{\sqrt{5}}{-1}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{5}}{2*-2}=\frac{0+4\sqrt{5}}{-4}
=\frac{4\sqrt{5}}{-4}
=\frac{\sqrt{5}}{-1}
$