10x^2+x-2=x(3x-2)

Solution for 10x^2+x-2=x(3x-2) equation:

10x^2+x-2=x(3x-2)

We move all terms to the left:

10x^2+x-2-(x(3x-2))=0


We calculate terms in parentheses: -(x(3x-2)), so:

x(3x-2)

We multiply parentheses

3x^2-2x

Back to the equation:

-(3x^2-2x)


We get rid of parentheses

10x^2-3x^2+x+2x-2=0

We add all the numbers together, and all the variables

7x^2+3x-2=0

a = 7; b = 3; c = -2;
Δ = b2-4ac
Δ = 32-4·7·(-2)
Δ = 65
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{65}}{2*7}=\frac{-3-\sqrt{65}}{14}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{65}}{2*7}=\frac{-3+\sqrt{65}}{14}
$