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

The solutions to the equation 4x^2-2x=23 are x=(1+sqrt(93))/4 ,x=(1-sqrt(93))/4

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

The zeros of 4x^2-2x=23 are x=(1+sqrt(93))/4 ,x=(1-sqrt(93))/4

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4 x^{2} - 2 x = 23

x = \frac{1 + 93}{4}, x = \frac{1 - 93}{4}

Decimal Notation

x = 2.66091 \dots, x = - 2.16091 \dots

Solution steps

4 x^{2} - 2 x = 23

Move

23

to the left side

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

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 4 (- 23) = 293

x_{1, 2} = \frac{- ( - 2 ) \pm 2 93}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 93}{2 \cdot 4}, x_{2} = \frac{- ( - 2 ) - 2 93}{2 \cdot 4}

x = \frac{- ( - 2 ) + 2 93}{2 \cdot 4} : \frac{1 + 93}{4}

x = \frac{- ( - 2 ) - 2 93}{2 \cdot 4} : \frac{1 - 93}{4}

The solutions to the quadratic equation are:

x = \frac{1 + 93}{4}, x = \frac{1 - 93}{4}

- x^{2} + 10 x - 24 = 0

- 12 + 5 (7 p + 3) - 28 p < 3 p - 17

46 + 9 a = - 5 a + 74

2 (x + 3) = 14

4 \leq x < 6

What are the solutions to the equation 4x^2-2x=23 ?
The solutions to the equation 4x^2-2x=23 are x=(1+sqrt(93))/4 ,x=(1-sqrt(93))/4
Find the zeros of 4x^2-2x=23
The zeros of 4x^2-2x=23 are x=(1+sqrt(93))/4 ,x=(1-sqrt(93))/4