What are the solutions to the equation 4x^2-23x-25=0 ?

The solutions to the equation 4x^2-23x-25=0 are x=(23+sqrt(929))/8 ,x=(23-sqrt(929))/8

Find the zeros of 4x^2-23x-25=0

The zeros of 4x^2-23x-25=0 are x=(23+sqrt(929))/8 ,x=(23-sqrt(929))/8

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4 x^{2} - 23 x - 25 = 0

x = \frac{23 + 929}{8}, x = \frac{23 - 929}{8}

Decimal Notation

x = 6.68493 \dots, x = - 0.93493 \dots

Solution steps

4 x^{2} - 23 x - 25 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 23 ) \pm ( - 23 ) ^{2} - 4 \cdot 4 ( - 25 )}{2 \cdot 4}

(- 23)^{2} - 4 \cdot 4 (- 25) = 929

x_{1, 2} = \frac{- ( - 23 ) \pm 929}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 23 ) + 929}{2 \cdot 4}, x_{2} = \frac{- ( - 23 ) - 929}{2 \cdot 4}

x = \frac{- ( - 23 ) + 929}{2 \cdot 4} : \frac{23 + 929}{8}

x = \frac{- ( - 23 ) - 929}{2 \cdot 4} : \frac{23 - 929}{8}

The solutions to the quadratic equation are:

x = \frac{23 + 929}{8}, x = \frac{23 - 929}{8}

4 n^{2} + 4 n - 8 = 0

∣ x - 5 ∣ < - 5

∣ 4 x - 5 ∣ - 2 = x + 1

- 4 x - 7 \leq 17

1 < 3 x + 7 < 19

What are the solutions to the equation 4x^2-23x-25=0 ?
The solutions to the equation 4x^2-23x-25=0 are x=(23+sqrt(929))/8 ,x=(23-sqrt(929))/8
Find the zeros of 4x^2-23x-25=0
The zeros of 4x^2-23x-25=0 are x=(23+sqrt(929))/8 ,x=(23-sqrt(929))/8