(x^2-3x+6)=(3x^2+5x-4)

Solution for (x^2-3x+6)=(3x^2+5x-4) equation:

(x^2-3x+6)=(3x^2+5x-4)

We move all terms to the left:

(x^2-3x+6)-((3x^2+5x-4))=0

We get rid of parentheses

x^2-3x-((3x^2+5x-4))+6=0


We calculate terms in parentheses: -((3x^2+5x-4)), so:

(3x^2+5x-4)

We get rid of parentheses

3x^2+5x-4

Back to the equation:

-(3x^2+5x-4)


We get rid of parentheses

x^2-3x^2-3x-5x+4+6=0

We add all the numbers together, and all the variables

-2x^2-8x+10=0

a = -2; b = -8; c = +10;
Δ = b2-4ac
Δ = -82-4·(-2)·10
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-12}{2*-2}=\frac{-4}{-4}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+12}{2*-2}=\frac{20}{-4}
=-5
$