4=1.28w^2/3

Solution for 4=1.28w^2/3 equation:

4=1.28w^2/3

We move all terms to the left:

4-(1.28w^2/3)=0

We get rid of parentheses

-1.28w^2/3+4=0

We multiply all the terms by the denominator


-1.28w^2+4*3=0

We add all the numbers together, and all the variables

-1.28w^2+12=0

a = -1.28; b = 0; c = +12;
Δ = b2-4ac
Δ = 02-4·(-1.28)·12
Δ = 61.44
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-\sqrt{61.44}}{2*-1.28}=\frac{0-\sqrt{61.44}}{-2.56}
=-\frac{\sqrt{}}{-2.56}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{61.44}}{2*-1.28}=\frac{0+\sqrt{61.44}}{-2.56}
=\frac{\sqrt{}}{-2.56}
$