What are the solutions to the equation 2x^2-3x+4=x+3 ?

The solutions to the equation 2x^2-3x+4=x+3 are x=(2+sqrt(2))/2 ,x=(2-sqrt(2))/2

Find the zeros of 2x^2-3x+4=x+3

The zeros of 2x^2-3x+4=x+3 are x=(2+sqrt(2))/2 ,x=(2-sqrt(2))/2

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2 x^{2} - 3 x + 4 = x + 3

x = \frac{2 + 2}{2}, x = \frac{2 - 2}{2}

Solution steps

2 x^{2} - 3 x + 4 = x + 3

Move

3

to the left side

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

Move

x

to the left side

2 x^{2} - 4 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 2 \cdot 1}{2 \cdot 2}

(- 4)^{2} - 4 \cdot 2 \cdot 1 = 22

x_{1, 2} = \frac{- ( - 4 ) \pm 2 2}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 2}{2 \cdot 2}, x_{2} = \frac{- ( - 4 ) - 2 2}{2 \cdot 2}

x = \frac{- ( - 4 ) + 2 2}{2 \cdot 2} : \frac{2 + 2}{2}

x = \frac{- ( - 4 ) - 2 2}{2 \cdot 2} : \frac{2 - 2}{2}

The solutions to the quadratic equation are:

x = \frac{2 + 2}{2}, x = \frac{2 - 2}{2}

\frac{∣ 2 - x ∣ - x ^{2}}{8 x - ∣ 9 - x ^{2} ∣} \leq 0

14 - (9 t - 2) < - t + 30

- 4 (\frac{588}{12881})

6 - \frac{1}{3} x^{2} = - 20

y^{2} - 9 y + 4 = 0

What are the solutions to the equation 2x^2-3x+4=x+3 ?
The solutions to the equation 2x^2-3x+4=x+3 are x=(2+sqrt(2))/2 ,x=(2-sqrt(2))/2
Find the zeros of 2x^2-3x+4=x+3
The zeros of 2x^2-3x+4=x+3 are x=(2+sqrt(2))/2 ,x=(2-sqrt(2))/2