What are the solutions to the equation x^2-4x-9=29 ?

The solutions to the equation x^2-4x-9=29 are x=2+sqrt(42),x=2-sqrt(42)

Find the zeros of x^2-4x-9=29

The zeros of x^2-4x-9=29 are x=2+sqrt(42),x=2-sqrt(42)

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x^{2} - 4 x - 9 = 29

x = 2 + 42, x = 2 - 42

Solution steps

x^{2} - 4 x - 9 = 29

Move

29

to the left side

x^{2} - 4 x - 38 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot ( - 38 )}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot (- 38) = 242

x_{1, 2} = \frac{- ( - 4 ) \pm 2 42}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 42}{2 \cdot 1}, x_{2} = \frac{- ( - 4 ) - 2 42}{2 \cdot 1}

x = \frac{- ( - 4 ) + 2 42}{2 \cdot 1} : 2 + 42

x = \frac{- ( - 4 ) - 2 42}{2 \cdot 1} : 2 - 42

The solutions to the quadratic equation are:

x = 2 + 42, x = 2 - 42

\frac{( x - 9 ) ( x + 1 )}{( x - 2 )} \geq 0

(y^{9})^{- 2}

9 (x - 4) = - 36

1^{2}

(27)^{\frac{- 2}{3}}

What are the solutions to the equation x^2-4x-9=29 ?
The solutions to the equation x^2-4x-9=29 are x=2+sqrt(42),x=2-sqrt(42)
Find the zeros of x^2-4x-9=29
The zeros of x^2-4x-9=29 are x=2+sqrt(42),x=2-sqrt(42)