What are the solutions to the equation x^2-2x-28=0 ?

The solutions to the equation x^2-2x-28=0 are x=1+sqrt(29),x=1-sqrt(29)

Find the zeros of x^2-2x-28=0

The zeros of x^2-2x-28=0 are x=1+sqrt(29),x=1-sqrt(29)

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x^{2} - 2 x - 28 = 0

x = 1 + 29, x = 1 - 29

Solution steps

x^{2} - 2 x - 28 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 28 )}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot (- 28) = 229

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 1 + 29, x = 1 - 29

∣ 5 - 2 x ∣ \leq 7

3 z^{2} + 10 z + 3

∣ x + 4 ∣ = 8

4 x^{2} + 21 x + 5 = 0

\frac{1}{x + 3} > \frac{1}{4}

What are the solutions to the equation x^2-2x-28=0 ?
The solutions to the equation x^2-2x-28=0 are x=1+sqrt(29),x=1-sqrt(29)
Find the zeros of x^2-2x-28=0
The zeros of x^2-2x-28=0 are x=1+sqrt(29),x=1-sqrt(29)