4x^2/3+8=40

Solution for 4x^2/3+8=40 equation:

4x^2/3+8=40

We move all terms to the left:

4x^2/3+8-(40)=0

We add all the numbers together, and all the variables

4x^2/3-32=0

We multiply all the terms by the denominator


4x^2-32*3=0

We add all the numbers together, and all the variables

4x^2-96=0

a = 4; b = 0; c = -96;
Δ = b2-4ac
Δ = 02-4·4·(-96)
Δ = 1536
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1536}=\sqrt{256*6}=\sqrt{256}*\sqrt{6}=16\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{6}}{2*4}=\frac{0-16\sqrt{6}}{8}
=-\frac{16\sqrt{6}}{8}
=-2\sqrt{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{6}}{2*4}=\frac{0+16\sqrt{6}}{8}
=\frac{16\sqrt{6}}{8}
=2\sqrt{6}
$