What are the solutions to the equation 2x^2-7x=5 ?

The solutions to the equation 2x^2-7x=5 are x=(7+sqrt(89))/4 ,x=(7-sqrt(89))/4

Find the zeros of 2x^2-7x=5

The zeros of 2x^2-7x=5 are x=(7+sqrt(89))/4 ,x=(7-sqrt(89))/4

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2 x^{2} - 7 x = 5

x = \frac{7 + 89}{4}, x = \frac{7 - 89}{4}

Decimal Notation

x = 4.10849 \dots, x = - 0.60849 \dots

Solution steps

2 x^{2} - 7 x = 5

Move

5

to the left side

2 x^{2} - 7 x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 2 ( - 5 )}{2 \cdot 2}

(- 7)^{2} - 4 \cdot 2 (- 5) = 89

x_{1, 2} = \frac{- ( - 7 ) \pm 89}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 89}{2 \cdot 2}, x_{2} = \frac{- ( - 7 ) - 89}{2 \cdot 2}

x = \frac{- ( - 7 ) + 89}{2 \cdot 2} : \frac{7 + 89}{4}

x = \frac{- ( - 7 ) - 89}{2 \cdot 2} : \frac{7 - 89}{4}

The solutions to the quadratic equation are:

x = \frac{7 + 89}{4}, x = \frac{7 - 89}{4}

\frac{5}{3} + 6

2 x^{2} - 9 x - 56

3 x^{2} - 4 x + 7 = 0

- 4 + x + 7 x^{2} = 0

\frac{1}{100} + \frac{1}{200} + \frac{1}{300}

What are the solutions to the equation 2x^2-7x=5 ?
The solutions to the equation 2x^2-7x=5 are x=(7+sqrt(89))/4 ,x=(7-sqrt(89))/4
Find the zeros of 2x^2-7x=5
The zeros of 2x^2-7x=5 are x=(7+sqrt(89))/4 ,x=(7-sqrt(89))/4