-2(x^2-3)=x^2-42

Solution for -2(x^2-3)=x^2-42 equation:

-2(x^2-3)=x^2-42

We move all terms to the left:

-2(x^2-3)-(x^2-42)=0

We multiply parentheses

-2x^2-(x^2-42)+6=0

We get rid of parentheses

-2x^2-x^2+42+6=0

We add all the numbers together, and all the variables

-3x^2+48=0

a = -3; b = 0; c = +48;
Δ = b2-4ac
Δ = 02-4·(-3)·48
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24}{2*-3}=\frac{-24}{-6}
=+4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24}{2*-3}=\frac{24}{-6}
=-4
$