What are the solutions to the equation x^2-4x-18=0 ?

The solutions to the equation x^2-4x-18=0 are x=2+sqrt(22),x=2-sqrt(22)

Find the zeros of x^2-4x-18=0

The zeros of x^2-4x-18=0 are x=2+sqrt(22),x=2-sqrt(22)

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x^{2} - 4 x - 18 = 0

x = 2 + 22, x = 2 - 22

Decimal Notation

x = 6.69041 \dots, x = - 2.69041 \dots

Solution steps

x^{2} - 4 x - 18 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot ( - 18 )}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot (- 18) = 222

x_{1, 2} = \frac{- ( - 4 ) \pm 2 22}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 22}{2 \cdot 1}, x_{2} = \frac{- ( - 4 ) - 2 22}{2 \cdot 1}

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

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

The solutions to the quadratic equation are:

x = 2 + 22, x = 2 - 22

1 lo g_{3} (3)

- x^{2} + 4 x - 7 = 0

x - (- 64) = - 18 + 27

- 13 = 5 (1 + 4 m) - 2 m

(3 - 2) (3 + 2)

What are the solutions to the equation x^2-4x-18=0 ?
The solutions to the equation x^2-4x-18=0 are x=2+sqrt(22),x=2-sqrt(22)
Find the zeros of x^2-4x-18=0
The zeros of x^2-4x-18=0 are x=2+sqrt(22),x=2-sqrt(22)