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

6 0

3 years ago

36/6=<br><br>I need help ASAP

egoroff_w [7]

Answer:

What?

Step-by-step explanation:

Are you trying to divide?

7 0

3 years ago

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Directions in picture.

densk [106]

Use PEMDAS, the answer that I got is 19.

8 0

2 years ago

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Solve by substitution <br> X^2= 2y + 10<br> 3x-y = 9

Viefleur [7K]

Answer:

$x=-4\\ x=-2\\ y=-21\\ y=-15$

Step-by-step explanation:

$x^{2} =2y+10$

$y=-9+3x$

$x^{2} =2(-9+3x)+10$

$x^{2} =-18+6x+10$

$x^{2} =6x-8$

$x^{2} -6x+8$

$x=-4, -2$

$y=-9+3(-4)$

$y=-9-12$

$y=-21$

$y=-9+3(-2)$

$y=-9-6$

$y=-15$

8 0

3 years ago

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