2(5x-12)=(3x-10)+(x^2-2)

Solution for 2(5x-12)=(3x-10)+(x^2-2) equation:

2(5x-12)=(3x-10)+(x^2-2)

We move all terms to the left:

2(5x-12)-((3x-10)+(x^2-2))=0

We multiply parentheses

10x-((3x-10)+(x^2-2))-24=0


We calculate terms in parentheses: -((3x-10)+(x^2-2)), so:

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

We get rid of parentheses

x^2+3x-10-2

We add all the numbers together, and all the variables

x^2+3x-12

Back to the equation:

-(x^2+3x-12)


We get rid of parentheses

-x^2+10x-3x+12-24=0

We add all the numbers together, and all the variables

-1x^2+7x-12=0

a = -1; b = 7; c = -12;
Δ = b2-4ac
Δ = 72-4·(-1)·(-12)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-1}{2*-1}=\frac{-8}{-2}
=+4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+1}{2*-1}=\frac{-6}{-2}
=+3
$