What are the solutions to the equation x^2-6x+13=32 ?

The solutions to the equation x^2-6x+13=32 are x=3+2sqrt(7),x=3-2sqrt(7)

Find the zeros of x^2-6x+13=32

The zeros of x^2-6x+13=32 are x=3+2sqrt(7),x=3-2sqrt(7)

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x^{2} - 6 x + 13 = 32

x = 3 + 27, x = 3 - 27

Decimal Notation

x = 8.29150 \dots, x = - 2.29150 \dots

Solution steps

x^{2} - 6 x + 13 = 32

Move

32

to the left side

x^{2} - 6 x - 19 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot ( - 19 )}{2 \cdot 1}

(- 6)^{2} - 4 \cdot 1 \cdot (- 19) = 47

x_{1, 2} = \frac{- ( - 6 ) \pm 4 7}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 6 ) + 4 7}{2 \cdot 1}, x_{2} = \frac{- ( - 6 ) - 4 7}{2 \cdot 1}

x = \frac{- ( - 6 ) + 4 7}{2 \cdot 1} : 3 + 27

x = \frac{- ( - 6 ) - 4 7}{2 \cdot 1} : 3 - 27

The solutions to the quadratic equation are:

x = 3 + 27, x = 3 - 27

\frac{9 a ^{4} b ^{3} - 3 a ^{3} b ^{4}}{6 a ^{3} b ^{3} - 2 a ^{2} b ^{4}}

\frac{2}{x + 2} > \frac{1}{x + 3}

2 (3 x - 4) \leq - 20

- 16 x^{2} - 6 x^{4}

- 3 (5 y - 4) \geq 17

What are the solutions to the equation x^2-6x+13=32 ?
The solutions to the equation x^2-6x+13=32 are x=3+2sqrt(7),x=3-2sqrt(7)
Find the zeros of x^2-6x+13=32
The zeros of x^2-6x+13=32 are x=3+2sqrt(7),x=3-2sqrt(7)