2.5z(2)=440

Solution for 2.5z(2)=440 equation:

2.5z(2)=440

We move all terms to the left:

2.5z(2)-(440)=0

We add all the numbers together, and all the variables

2.5z^2-440=0

a = 2.5; b = 0; c = -440;
Δ = b2-4ac
Δ = 02-4·2.5·(-440)
Δ = 4400
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:

$\sqrt{\Delta}=\sqrt{4400}=\sqrt{400*11}=\sqrt{400}*\sqrt{11}=20\sqrt{11}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{11}}{2*2.5}=\frac{0-20\sqrt{11}}{5}
=-\frac{20\sqrt{11}}{5}
=-4\sqrt{11}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{11}}{2*2.5}=\frac{0+20\sqrt{11}}{5}
=\frac{20\sqrt{11}}{5}
=4\sqrt{11}
$