3/2z-40=7z

Solution for 3/2z-40=7z equation:

3/2z-40=7z

We move all terms to the left:

3/2z-40-(7z)=0


Domain of the equation: 2z!=0

z!=0/2

z!=0

z∈R


We add all the numbers together, and all the variables

-7z+3/2z-40=0

We multiply all the terms by the denominator


-7z*2z-40*2z+3=0

Wy multiply elements

-14z^2-80z+3=0

a = -14; b = -80; c = +3;
Δ = b2-4ac
Δ = -802-4·(-14)·3
Δ = 6568
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:

$\sqrt{\Delta}=\sqrt{6568}=\sqrt{4*1642}=\sqrt{4}*\sqrt{1642}=2\sqrt{1642}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-80)-2\sqrt{1642}}{2*-14}=\frac{80-2\sqrt{1642}}{-28}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-80)+2\sqrt{1642}}{2*-14}=\frac{80+2\sqrt{1642}}{-28}
$