2z^2+3z=5-2z/4+z

Solution for 2z^2+3z=5-2z/4+z equation:

2z^2+3z=5-2z/4+z

We move all terms to the left:

2z^2+3z-(5-2z/4+z)=0


Domain of the equation: 4+z)!=0

We move all terms containing z to the left, all other terms to the right

z)!=-4

z!=-4/1

z!=-4

z∈R


We add all the numbers together, and all the variables

2z^2+3z-(z-2z/4+5)=0

We get rid of parentheses

2z^2+3z-z+2z/4-5=0

We multiply all the terms by the denominator


2z^2*4+3z*4-z*4+2z-5*4=0

We add all the numbers together, and all the variables

2z^2*4+2z+3z*4-z*4-20=0

Wy multiply elements

8z^2+2z+12z-4z-20=0

We add all the numbers together, and all the variables

8z^2+10z-20=0

a = 8; b = 10; c = -20;
Δ = b2-4ac
Δ = 102-4·8·(-20)
Δ = 740
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:

$\sqrt{\Delta}=\sqrt{740}=\sqrt{4*185}=\sqrt{4}*\sqrt{185}=2\sqrt{185}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{185}}{2*8}=\frac{-10-2\sqrt{185}}{16}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{185}}{2*8}=\frac{-10+2\sqrt{185}}{16}
$