What are the solutions to the equation x^2-2x-3=4 ?

The solutions to the equation x^2-2x-3=4 are x=1+2sqrt(2),x=1-2sqrt(2)

Find the zeros of x^2-2x-3=4

The zeros of x^2-2x-3=4 are x=1+2sqrt(2),x=1-2sqrt(2)

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x^{2} - 2 x - 3 = 4

x = 1 + 22, x = 1 - 22

Solution steps

x^{2} - 2 x - 3 = 4

Move

4

to the left side

x^{2} - 2 x - 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 7 )}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot (- 7) = 42

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

Separate the solutions

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

x = \frac{- ( - 2 ) + 4 2}{2 \cdot 1} : 1 + 22

x = \frac{- ( - 2 ) - 4 2}{2 \cdot 1} : 1 - 22

The solutions to the quadratic equation are:

x = 1 + 22, x = 1 - 22

(6 k^{4}) (- k^{2})

\frac{19 - 4 i}{2 - 5 i} + \frac{3 + 2 i}{i}

6 x^{2} - 55 x + 9

25 a^{2} - 20 a + 4

3 x^{2} = 4 x - 6

What are the solutions to the equation x^2-2x-3=4 ?
The solutions to the equation x^2-2x-3=4 are x=1+2sqrt(2),x=1-2sqrt(2)
Find the zeros of x^2-2x-3=4
The zeros of x^2-2x-3=4 are x=1+2sqrt(2),x=1-2sqrt(2)