What are the solutions to the equation 2x^2-5x=-1 ?

The solutions to the equation 2x^2-5x=-1 are x=(5+sqrt(17))/4 ,x=(5-sqrt(17))/4

Find the zeros of 2x^2-5x=-1

The zeros of 2x^2-5x=-1 are x=(5+sqrt(17))/4 ,x=(5-sqrt(17))/4

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2 x^{2} - 5 x = - 1

x = \frac{5 + 17}{4}, x = \frac{5 - 17}{4}

Solution steps

2 x^{2} - 5 x = - 1

Move

1

to the left side

2 x^{2} - 5 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 2 \cdot 1}{2 \cdot 2}

(- 5)^{2} - 4 \cdot 2 \cdot 1 = 17

x_{1, 2} = \frac{- ( - 5 ) \pm 17}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 17}{2 \cdot 2}, x_{2} = \frac{- ( - 5 ) - 17}{2 \cdot 2}

x = \frac{- ( - 5 ) + 17}{2 \cdot 2} : \frac{5 + 17}{4}

x = \frac{- ( - 5 ) - 17}{2 \cdot 2} : \frac{5 - 17}{4}

The solutions to the quadratic equation are:

x = \frac{5 + 17}{4}, x = \frac{5 - 17}{4}

2 x (x - \frac{7}{2}) = 49

r, r^{2} + 1 = 0

6 x^{2} + 24 x - 30 = 0

- 3 x^{2} + 14 x - 14 = 0

y, (x - 3)^{2} + (y + 2)^{2} = 9

What are the solutions to the equation 2x^2-5x=-1 ?
The solutions to the equation 2x^2-5x=-1 are x=(5+sqrt(17))/4 ,x=(5-sqrt(17))/4
Find the zeros of 2x^2-5x=-1
The zeros of 2x^2-5x=-1 are x=(5+sqrt(17))/4 ,x=(5-sqrt(17))/4