What are the solutions to the equation quadraticformula x^2+7x-4=0 ?

The solutions to the equation quadraticformula x^2+7x-4=0 are x=(-7+sqrt(65))/2 ,x=(-7-sqrt(65))/2

Find the zeros of quadraticformula x^2+7x-4=0

The zeros of quadraticformula x^2+7x-4=0 are x=(-7+sqrt(65))/2 ,x=(-7-sqrt(65))/2

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

x = \frac{- 7 + 65}{2}, x = \frac{- 7 - 65}{2}

Decimal Notation

x = 0.53112 \dots, x = - 7.53112 \dots

Solution steps

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

Solve with the quadratic formula

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

7^{2} - 4 \cdot 1 \cdot (- 4) = 65

x_{1, 2} = \frac{- 7 \pm 65}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 7 + 65}{2 \cdot 1}, x_{2} = \frac{- 7 - 65}{2 \cdot 1}

x = \frac{- 7 + 65}{2 \cdot 1} : \frac{- 7 + 65}{2}

x = \frac{- 7 - 65}{2 \cdot 1} : \frac{- 7 - 65}{2}

The solutions to the quadratic equation are:

x = \frac{- 7 + 65}{2}, x = \frac{- 7 - 65}{2}

x, x - \frac{5}{x} < 4

r, 5 (r - 1) = 2 (r - 4) - 6

32.5%

60 \cdot 25%

n, 2 - \frac{1}{2} n = 3 n + 16

What are the solutions to the equation quadraticformula x^2+7x-4=0 ?
The solutions to the equation quadraticformula x^2+7x-4=0 are x=(-7+sqrt(65))/2 ,x=(-7-sqrt(65))/2
Find the zeros of quadraticformula x^2+7x-4=0
The zeros of quadraticformula x^2+7x-4=0 are x=(-7+sqrt(65))/2 ,x=(-7-sqrt(65))/2