(4/3)x^2=5x

Solution for (4/3)x^2=5x equation:

(4/3)x^2=5x

We move all terms to the left:

(4/3)x^2-(5x)=0


Domain of the equation: 3)x^2!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

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

We multiply all the terms by the denominator


-5x*3)x^2+(+4=0

Wy multiply elements

-15x^2+4=0

a = -15; b = 0; c = +4;
Δ = b2-4ac
Δ = 02-4·(-15)·4
Δ = 240
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:

$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{15}}{2*-15}=\frac{0-4\sqrt{15}}{-30}
=-\frac{4\sqrt{15}}{-30}
=-\frac{2\sqrt{15}}{-15}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{15}}{2*-15}=\frac{0+4\sqrt{15}}{-30}
=\frac{4\sqrt{15}}{-30}
=\frac{2\sqrt{15}}{-15}
$