What are the solutions to the equation x^2+5x-4=0 ?

The solutions to the equation x^2+5x-4=0 are x=(-5+sqrt(41))/2 ,x=(-5-sqrt(41))/2

Find the zeros of x^2+5x-4=0

The zeros of x^2+5x-4=0 are x=(-5+sqrt(41))/2 ,x=(-5-sqrt(41))/2

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x^{2} + 5 x - 4 = 0

x = \frac{- 5 + 41}{2}, x = \frac{- 5 - 41}{2}

Decimal Notation

x = 0.70156 \dots, x = - 5.70156 \dots

Solution steps

x^{2} + 5 x - 4 = 0

Solve with the quadratic formula

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

5^{2} - 4 \cdot 1 \cdot (- 4) = 41

x_{1, 2} = \frac{- 5 \pm 41}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 5 + 41}{2 \cdot 1}, x_{2} = \frac{- 5 - 41}{2 \cdot 1}

x = \frac{- 5 + 41}{2 \cdot 1} : \frac{- 5 + 41}{2}

x = \frac{- 5 - 41}{2 \cdot 1} : \frac{- 5 - 41}{2}

The solutions to the quadratic equation are:

x = \frac{- 5 + 41}{2}, x = \frac{- 5 - 41}{2}

\frac{1}{2} (x - 1) - \frac{2}{3} (x + 1) = 1

6 - 2 (- 2 h + 1) \leq 5 h + 5 + h

\frac{1}{2} t^{2} - 25 = 0

2 y^{4} - y^{3} + 4 y - 2

\frac{2}{5} x + 7 = \frac{5}{2} x + 3

What are the solutions to the equation x^2+5x-4=0 ?
The solutions to the equation x^2+5x-4=0 are x=(-5+sqrt(41))/2 ,x=(-5-sqrt(41))/2
Find the zeros of x^2+5x-4=0
The zeros of x^2+5x-4=0 are x=(-5+sqrt(41))/2 ,x=(-5-sqrt(41))/2