What are the solutions to the equation x^2-5x-25=4 ?

The solutions to the equation x^2-5x-25=4 are x=(5+sqrt(141))/2 ,x=(5-sqrt(141))/2

Find the zeros of x^2-5x-25=4

The zeros of x^2-5x-25=4 are x=(5+sqrt(141))/2 ,x=(5-sqrt(141))/2

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x^{2} - 5 x - 25 = 4

x = \frac{5 + 141}{2}, x = \frac{5 - 141}{2}

Decimal Notation

x = 8.43717 \dots, x = - 3.43717 \dots

Solution steps

x^{2} - 5 x - 25 = 4

Move

4

to the left side

x^{2} - 5 x - 29 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot ( - 29 )}{2 \cdot 1}

(- 5)^{2} - 4 \cdot 1 \cdot (- 29) = 141

x_{1, 2} = \frac{- ( - 5 ) \pm 141}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 141}{2 \cdot 1}, x_{2} = \frac{- ( - 5 ) - 141}{2 \cdot 1}

x = \frac{- ( - 5 ) + 141}{2 \cdot 1} : \frac{5 + 141}{2}

x = \frac{- ( - 5 ) - 141}{2 \cdot 1} : \frac{5 - 141}{2}

The solutions to the quadratic equation are:

x = \frac{5 + 141}{2}, x = \frac{5 - 141}{2}

4 x + 12 = x^{2}

x^{2} - 5 = x + 5

x, x^{2} + 2 y^{2} + z = 16

- 3 x^{2} - x - 1 = 0

\frac{9 x ^{2}}{5} - \frac{7 x}{3} + 2 = 2 x^{2} - \frac{x}{2} + \frac{1}{3}

What are the solutions to the equation x^2-5x-25=4 ?
The solutions to the equation x^2-5x-25=4 are x=(5+sqrt(141))/2 ,x=(5-sqrt(141))/2
Find the zeros of x^2-5x-25=4
The zeros of x^2-5x-25=4 are x=(5+sqrt(141))/2 ,x=(5-sqrt(141))/2