40/2z+3+5-3z;z=-3*

Solution for 40/2z+3+5-3z;z=-3* equation:

40/2z+3+5-3zz=-3*

We move all terms to the left:

40/2z+3+5-3zz-(-3*)=0


Domain of the equation: 2z!=0

z!=0/2

z!=0

z∈R


We add all the numbers together, and all the variables

40/2z-3zz+3+5-0=0

We add all the numbers together, and all the variables

40/2z-3zz+8=0

We multiply all the terms by the denominator


-3zz*2z+8*2z+40=0

Wy multiply elements

-6z^2+16z+40=0

a = -6; b = 16; c = +40;
Δ = b2-4ac
Δ = 162-4·(-6)·40
Δ = 1216
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:

$\sqrt{\Delta}=\sqrt{1216}=\sqrt{64*19}=\sqrt{64}*\sqrt{19}=8\sqrt{19}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8\sqrt{19}}{2*-6}=\frac{-16-8\sqrt{19}}{-12}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8\sqrt{19}}{2*-6}=\frac{-16+8\sqrt{19}}{-12}
$