(-4/3z+41)+z=57

Solution for (-4/3z+41)+z=57 equation:

(-4/3z+41)+z=57

We move all terms to the left:

(-4/3z+41)+z-(57)=0


Domain of the equation: 3z+41)!=0

z∈R


We add all the numbers together, and all the variables

z+(-4/3z+41)-57=0

We get rid of parentheses

z-4/3z+41-57=0

We multiply all the terms by the denominator


z*3z+41*3z-57*3z-4=0

Wy multiply elements

3z^2+123z-171z-4=0

We add all the numbers together, and all the variables

3z^2-48z-4=0

a = 3; b = -48; c = -4;
Δ = b2-4ac
Δ = -482-4·3·(-4)
Δ = 2352
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:

$\sqrt{\Delta}=\sqrt{2352}=\sqrt{784*3}=\sqrt{784}*\sqrt{3}=28\sqrt{3}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-28\sqrt{3}}{2*3}=\frac{48-28\sqrt{3}}{6}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+28\sqrt{3}}{2*3}=\frac{48+28\sqrt{3}}{6}
$