What are the solutions to the equation x^2-4x=19 ?

The solutions to the equation x^2-4x=19 are x=2+sqrt(23),x=2-sqrt(23)

Find the zeros of x^2-4x=19

The zeros of x^2-4x=19 are x=2+sqrt(23),x=2-sqrt(23)

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x^{2} - 4 x = 19

x = 2 + 23, x = 2 - 23

Decimal Notation

x = 6.79583 \dots, x = - 2.79583 \dots

Solution steps

x^{2} - 4 x = 19

Move

19

to the left side

x^{2} - 4 x - 19 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot ( - 19 )}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot (- 19) = 223

x_{1, 2} = \frac{- ( - 4 ) \pm 2 23}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 23}{2 \cdot 1}, x_{2} = \frac{- ( - 4 ) - 2 23}{2 \cdot 1}

x = \frac{- ( - 4 ) + 2 23}{2 \cdot 1} : 2 + 23

x = \frac{- ( - 4 ) - 2 23}{2 \cdot 1} : 2 - 23

The solutions to the quadratic equation are:

x = 2 + 23, x = 2 - 23

5 - 3 (x + 2) = x - 9

5 X - (X - 11) = 8

\frac{7 x}{3} < 2

3 x^{3} y^{2} + 3 x^{2} + 6 x y^{2} + 6

(1 + i)^{20}

What are the solutions to the equation x^2-4x=19 ?
The solutions to the equation x^2-4x=19 are x=2+sqrt(23),x=2-sqrt(23)
Find the zeros of x^2-4x=19
The zeros of x^2-4x=19 are x=2+sqrt(23),x=2-sqrt(23)