(3/2x+20)+20x=90

Solution for (3/2x+20)+20x=90 equation:

(3/2x+20)+20x=90

We move all terms to the left:

(3/2x+20)+20x-(90)=0


Domain of the equation: 2x+20)!=0

x∈R


We add all the numbers together, and all the variables

20x+(3/2x+20)-90=0

We get rid of parentheses

20x+3/2x+20-90=0

We multiply all the terms by the denominator


20x*2x+20*2x-90*2x+3=0

Wy multiply elements

40x^2+40x-180x+3=0

We add all the numbers together, and all the variables

40x^2-140x+3=0

a = 40; b = -140; c = +3;
Δ = b2-4ac
Δ = -1402-4·40·3
Δ = 19120
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:

$\sqrt{\Delta}=\sqrt{19120}=\sqrt{16*1195}=\sqrt{16}*\sqrt{1195}=4\sqrt{1195}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-140)-4\sqrt{1195}}{2*40}=\frac{140-4\sqrt{1195}}{80}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-140)+4\sqrt{1195}}{2*40}=\frac{140+4\sqrt{1195}}{80}
$