What are the solutions to the equation x^2-4x=-1 ?

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

Find the zeros of x^2-4x=-1

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

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x^{2} - 4 x = - 1

x = 2 + 3, x = 2 - 3

Decimal Notation

x = 3.73205 \dots, x = 0.26794 \dots

Solution steps

x^{2} - 4 x = - 1

Move

1

to the left side

x^{2} - 4 x + 1 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 1 \cdot 1 = 23

x_{1, 2} = \frac{- ( - 4 ) \pm 2 3}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 3}{2 \cdot 1}, x_{2} = \frac{- ( - 4 ) - 2 3}{2 \cdot 1}

x = \frac{- ( - 4 ) + 2 3}{2 \cdot 1} : 2 + 3

x = \frac{- ( - 4 ) - 2 3}{2 \cdot 1} : 2 - 3

The solutions to the quadratic equation are:

x = 2 + 3, x = 2 - 3

6 x^{2} + 9 x

∣ p ∣ \leq 12

15 + 8 i

6 b c^{0}

\frac{12 !}{7 !}

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