How do you write $2 x^{2} - 3 = - 6 x - 2$ $2x^2 - 3 = -6x - 2$ in standard form?

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Answer 1 · 2016-04-15T01:36:56

$2 x^{2} + 6 x - 1 = 0$ $2x^2 +6x-1=0$

Standard form of a quadratic equation looks like this:
$a x^{2} + b x + c = 0$ $ax^2 + bx + c = 0$

So basically, we need one side of the equation to be $0$ $0$ .

$2 x^{2} - 3 = - 6 x - 2$ $2x^2-3=-6x-2$

add everything to the left side
$2 x^{2} - 3 + 6 x + 2 = 0$ $2x^2-3+6x+2=0$

simplify
$2 x^{2} + 6 x - 1 = 0$ $2x^2 +6x-1=0$

usually you want $a$ $a$ (which is $2$ $2$ in this case) to be positive, and it is so we are done.

How do you write 2x^2 - 3 = -6x - 22x2−3=−6x−22x^2 - 3 = -6x - 2 in standard form?