What are the solutions to the equation 2x^2-3x=1 ?

The solutions to the equation 2x^2-3x=1 are x=(3+sqrt(17))/4 ,x=(3-sqrt(17))/4

Find the zeros of 2x^2-3x=1

The zeros of 2x^2-3x=1 are x=(3+sqrt(17))/4 ,x=(3-sqrt(17))/4

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2 x^{2} - 3 x = 1

x = \frac{3 + 17}{4}, x = \frac{3 - 17}{4}

Decimal Notation

x = 1.78077 \dots, x = - 0.28077 \dots

Solution steps

2 x^{2} - 3 x = 1

Move

1

to the left side

2 x^{2} - 3 x - 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 2 ( - 1 )}{2 \cdot 2}

(- 3)^{2} - 4 \cdot 2 (- 1) = 17

x_{1, 2} = \frac{- ( - 3 ) \pm 17}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 17}{2 \cdot 2}, x_{2} = \frac{- ( - 3 ) - 17}{2 \cdot 2}

x = \frac{- ( - 3 ) + 17}{2 \cdot 2} : \frac{3 + 17}{4}

x = \frac{- ( - 3 ) - 17}{2 \cdot 2} : \frac{3 - 17}{4}

The solutions to the quadratic equation are:

x = \frac{3 + 17}{4}, x = \frac{3 - 17}{4}

0 = 1 - x^{2}

x^{2} + 15 x - 2 \geq 7 x + 18

36 m^{4} - 16 m^{2}

900 t^{2} - 400

\frac{2 i}{2 i}

What are the solutions to the equation 2x^2-3x=1 ?
The solutions to the equation 2x^2-3x=1 are x=(3+sqrt(17))/4 ,x=(3-sqrt(17))/4
Find the zeros of 2x^2-3x=1
The zeros of 2x^2-3x=1 are x=(3+sqrt(17))/4 ,x=(3-sqrt(17))/4