Solve for [tex]\( x \)[/tex]:
[tex]\[ 3x = 6x - 2 \][/tex]



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[tex]$
-4\left(2 x^2-3\right)=-8 x^2+c
$[/tex]

In the given equation, [tex]$c$[/tex] is a constant. What is the value of [tex]$c$[/tex]?
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Answer :

To find the value of the constant [tex]\(c\)[/tex] in the given equation

[tex]\[ -4\left(2 x^2 - 3\right) = -8 x^2 + c, \][/tex]

we need to simplify and compare both sides of the equation step-by-step.

1. Distribute the [tex]\(-4\)[/tex] on the left side of the equation:

[tex]\[ -4 \left(2 x^2 - 3\right) = -4 \cdot 2 x^2 -4 \cdot (-3). \][/tex]

Simplifying the multiplication, we get:

[tex]\[ -8 x^2 + 12. \][/tex]

2. Rewrite the given equation with the simplified left side:

[tex]\[ -8 x^2 + 12 = -8 x^2 + c. \][/tex]

3. Compare the simplified left side expression with the right side of the equation:

We can see that the terms involving [tex]\(x^2\)[/tex] are identical on both sides. Therefore, what remains is to equate the constant terms on each side. That is:

[tex]\[ 12 = c. \][/tex]

Therefore, the value of [tex]\(c\)[/tex] is:

[tex]\[ \boxed{12} \][/tex]