What is 3x^2-5x+3>2x^2+x-4 ?

The solution to 3x^2-5x+3>2x^2+x-4 is x sqrt(2)+3

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3x^2-5x+3>2x^2+x-4

3 x^{2} - 5 x + 3 > 2 x^{2} + x - 4

x < - 2 + 3 or x > 2 + 3

Interval Notation

(- \infty, - 2 + 3) \cup (2 + 3, \infty)

Decimal Notation

x < 1.58578 \dots or x > 4.41421 \dots

Solution steps

3 x^{2} - 5 x + 3 > 2 x^{2} + x - 4

Rewrite in standard form

x^{2} - 6 x + 7 > 0

Complete the square

x^{2} - 6 x + 7 : (x - 3)^{2} - 2

(x - 3)^{2} - 2 > 0

Move

2

to the right side

(x - 3)^{2} > 2

For

u^{n} > a

, if

n

is even then

x - 3 < - 2 or x - 3 > 2

x - 3 < - 2 : x < - 2 + 3

x - 3 > 2 : x > 2 + 3

Combine the intervals

x < - 2 + 3 or x > 2 + 3

x^{\frac{1}{4}} \cdot x^{\frac{1}{2}}

r - 2 = 5

e^{1} + e^{- 1}

f = 2 - 6

(- \frac{1}{4})^{4}

What is 3x^2-5x+3>2x^2+x-4 ?
The solution to 3x^2-5x+3>2x^2+x-4 is x<-sqrt(2)+3\lor x>sqrt(2)+3