How do you solve $2 x^{2} - 7 x + 12 = 0$ $2x^2-7x+12=0$ by completing the square?

Question

How do you solve $2 x^{2} - 7 x + 12 = 0$ $2x^2-7x+12=0$ by completing the square?

1 Answer

Impact of this question

6081 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-16T06:17:39

$x = \frac{7}{4} \pm \frac{i \sqrt{- 47}}{4}$ $x=7/4+-(isqrt(-47))/4$

Explanation:

Given -

$2 x^{2} - 7 x + 12 = 0$ $2x^2-7x+12=0$

$2 x^{2} - 7 x = - 12$ $2x^2-7x=-12$
Divide both sides by $2$ $2$
$(cancel(2)x^2)/cancel(2)-7/2x=(-12)/2$
$x^2-7/2x=-6$
Divide $7/2$ by $2$ , square it and add it to both sides

$x^2-7/2*1/2*x +49/16=-6+49/16$
$x^2-7/4+49/16=-6+49/16$
$x^2-7/4+49/16=(-96+49)/16$
$(x-7/4)^2=(-47)/16$
$x-7/4=+-sqrt((-47)/16)$
$x=7/4+-(isqrt(-47))/4$

Science

Math

Humanities

... and beyond

How do you solve $2 x^{2} - 7 x + 12 = 0$ $2x^2-7x+12=0$ by completing the square?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve 2x^2-7x+12=02x2−7x+12=02x^2-7x+12=0 by completing the square?

1 Answer

Explanation:

Related questions

Impact of this question

How do you solve $2 x^{2} - 7 x + 12 = 0$ $2x^2-7x+12=0$ by completing the square?