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

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

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

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

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

x = 1 + 22, x = 1 - 22

Decimal Notation

x = 3.82842 \dots, x = - 1.82842 \dots

Solution steps

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

Switch sides

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

100 x^{2} - 140 x = - 28

- 5 x^{2} + 2 x + 11 = 0

2 x^{2} - 2 x + 15 = 0

837.2 = 0.68 \cdot 2.14 \cdot 100 \cdot x \cdot (9.5 - 0.4 \cdot x)

12 x^{2} - 48 x = - 39

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