How do you solve $2x^2 – 28x + 84 = 0$ using completing the square?

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Answer 1 · 2015-07-03T14:07:40

Logical first step is dividing everything by $2$

$->x^2-14x+42=0$

Take half the $x$ -coefficient and square that:
$x^2-14x+(-7)^2=x^2-14x+49=(x-7)^2$

If we write the original $42=49-7$ :
$->(x-7)^2-7=0->(x-7)^2=7$

So: $x-7=sqrt7->x=7+sqrt7$
Or: $x-7=-sqrt7->x=7-sqrt7$

Often written as $x_(1,2)=7+-sqrt7$

How do you solve 2x^2 – 28x + 84 = 02x^2 – 28x + 84 = 0 using completing the square?