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3 years ago

Please help me out, I really could use it

Mathematics

1 answer:

SVETLANKA909090 [29]3 years ago

6 0

Answer:

No real solutions

129

Step-by-step explanation:

The quadratic formula is

-b ± sqrt(b^2 -4ac)

----------------------------

3x^2 =2x-1

Lets get the equation in proper form

3x^2 -2x+1 = 2x-1-2x+1

3x^2 -2x+1 =0

a=3 b=-2 c=1

Lets substitute what we know

2 ± sqrt((-2)^2 -4(3)(1))

----------------------------

2(2)

-2 ± sqrt(4-12)

----------------------------

2(2)

-2 ± sqrt(-8)

----------------------------

No real solutions

The quadratic formula is

-b ± sqrt(b^2 -4ac)

----------------------------

2x^2 -10= 7x

Lets get the equation in proper form

2x^2 -7x-10 = 7-7x

2x^2 -7x-10 =0

a=2 b=-7 c=-10

Lets substitute what we know, we are only looking for what is inside the radical

(b^2 -4ac)

((-7)^2 -4(2)(-10))

(49 +80)

129

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The full question is attached.

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