What are the solutions to the equation 4x^2-28x+29=0 ?

The solutions to the equation 4x^2-28x+29=0 are x=(7+2sqrt(5))/2 ,x=(7-2sqrt(5))/2

Find the zeros of 4x^2-28x+29=0

The zeros of 4x^2-28x+29=0 are x=(7+2sqrt(5))/2 ,x=(7-2sqrt(5))/2

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4 x^{2} - 28 x + 29 = 0

x = \frac{7 + 2 5}{2}, x = \frac{7 - 2 5}{2}

Decimal Notation

x = 5.73606 \dots, x = 1.26393 \dots

Solution steps

4 x^{2} - 28 x + 29 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 28 ) \pm ( - 28 ) ^{2} - 4 \cdot 4 \cdot 29}{2 \cdot 4}

(- 28)^{2} - 4 \cdot 4 \cdot 29 = 85

x_{1, 2} = \frac{- ( - 28 ) \pm 8 5}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 28 ) + 8 5}{2 \cdot 4}, x_{2} = \frac{- ( - 28 ) - 8 5}{2 \cdot 4}

x = \frac{- ( - 28 ) + 8 5}{2 \cdot 4} : \frac{7 + 2 5}{2}

x = \frac{- ( - 28 ) - 8 5}{2 \cdot 4} : \frac{7 - 2 5}{2}

The solutions to the quadratic equation are:

x = \frac{7 + 2 5}{2}, x = \frac{7 - 2 5}{2}

- 0.00204 x^{2} + 0.04318 x - 0.21133 = 0

x^{2} + 19 x - 66 = 0

16 + 0.5 (30 - A^{1}) = 10 + 2.5 A^{2}

12 (6 y) - 48 (y - 4) = 5 y (y - 4)

- 2 x^{2} + 5 x + 6 = 0

What are the solutions to the equation 4x^2-28x+29=0 ?
The solutions to the equation 4x^2-28x+29=0 are x=(7+2sqrt(5))/2 ,x=(7-2sqrt(5))/2
Find the zeros of 4x^2-28x+29=0
The zeros of 4x^2-28x+29=0 are x=(7+2sqrt(5))/2 ,x=(7-2sqrt(5))/2