$4x^2-8x-12=0\\{}\qquad\qquad^{\div4\qquad\div4}\\x^2-2x-3=0\\\\x^2+x-3x-3=0\\\\x(x+1)-3(x+1)=0\\\\(x+1)(x-3)=0\\\\x+1=0\quad\vee\quad x-3=0\\\\x=-1\quad\vee\quad x=3$

6 0

4 years ago

Rewrite this standard form quadratic equation into vertex form by completing the square. Upload your work here to show your thin

Alex787 [66]

Solution :

Given the equation :

$y = x^2 - 6x+3$

$y=1(x^2 - 6x +n)+3$

$n = \left(\frac{b}{2}\right)^2$

$n = \left(\frac{6}{2}\right)^2$

n = 9

$y=1(x^2 - 6x +9-9)+3$

$y=1(x^2 - 6x +9)-9+3$

$y=1(x-3)^2- 6$

Therefore the vertex form of $y = x^2 - 6x+3$ is $y=1(x-3)^2- 6$ .

3 0

3 years ago

Which expression is not equivalent to 7(3+5)

irakobra [83]

Answer:

Any expression that doesn't equal 56... You didn't really provide any answer choices, so that's the best answer I have for you...

4 0

3 years ago

Ill mark brainlist plss help

mojhsa [17]

Answer:

w=20

Step-by-step explanation:

Solve for w by cross multiplying.

6 0

3 years ago

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