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

The solutions to the equation x^2-4x=25 are x=2+sqrt(29),x=2-sqrt(29)

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

The zeros of x^2-4x=25 are x=2+sqrt(29),x=2-sqrt(29)

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x^{2} - 4 x = 25

x = 2 + 29, x = 2 - 29

Decimal Notation

x = 7.38516 \dots, x = - 3.38516 \dots

Solution steps

x^{2} - 4 x = 25

Move

25

to the left side

x^{2} - 4 x - 25 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot ( - 25 )}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot (- 25) = 229

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

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 29}{2 \cdot 1}, x_{2} = \frac{- ( - 4 ) - 2 29}{2 \cdot 1}

x = \frac{- ( - 4 ) + 2 29}{2 \cdot 1} : 2 + 29

x = \frac{- ( - 4 ) - 2 29}{2 \cdot 1} : 2 - 29

The solutions to the quadratic equation are:

x = 2 + 29, x = 2 - 29

x^{(2)} - 25 = 0

- 3 (x + 5) > 3

\frac{20 !}{14 ! \cdot 6 !}

43 \cdot 2

16 x^{2} c + 8 x y d - 16 x^{2} d - 8 x yc

What are the solutions to the equation x^2-4x=25 ?
The solutions to the equation x^2-4x=25 are x=2+sqrt(29),x=2-sqrt(29)
Find the zeros of x^2-4x=25
The zeros of x^2-4x=25 are x=2+sqrt(29),x=2-sqrt(29)