How do you solve $2x^2 - 12x + 10 = 0$ ?

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Answer 1 · 2015-11-04T17:31:05

The two solutions are $x=1$ and $x=5$ .

First of all, we can divide both members by $2$ :

$(2x^2-12x+10)/2 = 0/2$

which simplifies into

$x^2-6x+5=0$

Arrived at this point, we can complete the square:

$x^2-6x+9-9+5=0$

And we notice that $x^2-6x+9=(x-3)^2$ .

So, the whole expression becomes

$(x-3)^2 -4 = 0$

$\iff$

$(x-3)^2 = 4$

$\iff$

$x-3 = \pm 2$

$iff$

$x=3\pm 2 \implies x=1$ or $x=5$ .

How do you solve 2x^2 - 12x + 10 = 02x^2 - 12x + 10 = 0?