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

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

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

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

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

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

Decimal Notation

x = - 1.26393 \dots, x = - 5.73606 \dots

Solution steps

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

Move

28 x

to the left side

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

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 28 \pm 2 8 ^{2} - 4 \cdot 4 \cdot 29}{2 \cdot 4}

2 8^{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}

(x - 5)^{2} - (x - 6)^{2} = (2 x - 3)^{2} - 118

x^{2} = - 6 x - 9

x, f = x^{2} y

x (5 x - 1) = 6

x, y = x^{2}

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