L1=3x-2/1+2x^2

Solution for L1=3x-2/1+2x^2 equation:

1=3L-2/1+2L^2

We move all terms to the left:

1-(3L-2/1+2L^2)=0


Domain of the equation: 1+2L^2)!=0

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

2L^2)!=-1

L!=-1/1

L!=-1

L∈R


We get rid of parentheses

-2L^2-3L+1+2/1=0

We multiply all the terms by the denominator


-2L^2*1-3L*1+2+1*1=0

We add all the numbers together, and all the variables

-2L^2*1-3L*1+3=0

Wy multiply elements

-2L^2-3L+3=0

a = -2; b = -3; c = +3;
Δ = b2-4ac
Δ = -32-4·(-2)·3
Δ = 33
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$L_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$L_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$L_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{33}}{2*-2}=\frac{3-\sqrt{33}}{-4}
$
$L_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{33}}{2*-2}=\frac{3+\sqrt{33}}{-4}
$