Solve the quadratic equations:

(x-4)^2=20

x^2+4x-8=0

4x^2-6x+1=0

3x^2=6x+7

Can someone show me how to do these?

Question

Solve the quadratic equations:

(x-4)^2=20

x^2+4x-8=0

4x^2-6x+1=0

3x^2=6x+7

Can someone show me how to do these?

anonymous · Answer

(x-4)(x-4)=20
foil the left

mertsj · Answer

Take the square root of  both sides.

mertsj · Answer

x-4= +- sqrt20

mertsj · Answer

x=4+-sqrt20=4+-2sqrt5

mertsj · Answer

I just showed you the first one.

anonymous · Answer

once you foil it. then subtract the 20 from both sides

mertsj · Answer

Use quadratic formula on second one.

anonymous · Answer

Im stuck on the first one. None  of my answer choices is the answer you gave me.

mertsj · Answer

Also use the formula for 3 and 4

mertsj · Answer

What are your answer choices?

mertsj · Answer

Are they in decimal form or what?

anonymous · Answer

use the quadratic formula.. 

a(x^2) + bx + c = 0

a,b,c are constants

$$-b \pm \sqrt{b ^{2}-4ac}\div2a$$

anonymous · Answer

answer choices for number one:
a. x=4+- 2 SR5
b. x=+- 6 SR 5
c. x=4 +- 5 SR 2
d. x=14
e. x= +- 2 SR 6

mertsj · Answer

What answers do you think I posted for the first one?  Did you even read what I wrote?

mertsj · Answer

99
Mertsj






x=4+-sqrt20=4+-2sqrt5

anonymous · Answer

You know, yahl can band me from this site if yahl want, but your seriously a feather. You have the worst attitude EVER!

anonymous · Answer

* B   I   T   C   H, not feather.