2x(11/2-x)=213=1/15

Solution for 2x(11/2-x)=213=1/15 equation:

2x(11/2-x)=213=1/15

We move all terms to the left:

2x(11/2-x)-(213)=0


Domain of the equation: 2-x)!=0

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

-x)!=-2

x!=-2/1

x!=-2

x∈R


We add all the numbers together, and all the variables

2x(-1x+11/2)-213=0

We multiply parentheses

-2x^2+22x^2-213=0

We add all the numbers together, and all the variables

20x^2-213=0

a = 20; b = 0; c = -213;
Δ = b2-4ac
Δ = 02-4·20·(-213)
Δ = 17040
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:

$\sqrt{\Delta}=\sqrt{17040}=\sqrt{16*1065}=\sqrt{16}*\sqrt{1065}=4\sqrt{1065}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{1065}}{2*20}=\frac{0-4\sqrt{1065}}{40}
=-\frac{4\sqrt{1065}}{40}
=-\frac{\sqrt{1065}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{1065}}{2*20}=\frac{0+4\sqrt{1065}}{40}
=\frac{4\sqrt{1065}}{40}
=\frac{\sqrt{1065}}{10}
$