rewrite in standard form -2x^2+3x+7

## rewrite in standard form -2x^2+3x+7

# 1 Answer

Recall that the standard form of a quadratic equation is as follows: a(x - h)^{2} + k

-2x^{2} + 3x + 7 separate the first 2 terms from the last term by parentheses

(-2x^{2} + 3x) + 7

factor out the coefficient of the 1st term (a) from the first 2 terms which are inside the parentheses

-2(x^{2} - (3/2)x) + 7

Complete the square inside the parentheses:

x^{2} - (3/2)x ==> (b/2)^{2} = ((-3/2)/2)^{2} = (-3/4)^{2} = 9/16

Add and subtract the term that completes the square inside the parentheses then multiply the term that is subtracted by a (i.e., -2), which was factored out in the beginning, then move this term to the outside.

-2(x^{2} - (3/2)x + (9/16) - (9/16)) + 7

-2(x^{2} - (3/2)x + (9/16)) + 7 - (9/16)·(-2)

-2(x^{2} - (3/2)x + (9/16)) + 7 - (-2·9/16)

-2(x^{2} - (3/2)x + (9/16)) +
**7 + (18/16)**

==> **7 + (18/16)** = (7·16/16) + (18/16) = (112/16) + (18/16) = (112+18)/16 = 130/16 =
**65/8**

-2(x^{2} - (3/2)x + (9/16)) + 65/8

Since we completed the square for the equation inside the parentheses in the beginning, the equation inside the parentheses is now a perfect square:

-2(x - (3/4))(x - (3/4)) + 65/8

** -2(x - (3/4))**^{2}** + 65/8**