What are the solutions to the equation x^2-17x+13=-27 ?

The solutions to the equation x^2-17x+13=-27 are x=(17+sqrt(129))/2 ,x=(17-sqrt(129))/2

Find the zeros of x^2-17x+13=-27

The zeros of x^2-17x+13=-27 are x=(17+sqrt(129))/2 ,x=(17-sqrt(129))/2

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x^{2} - 17 x + 13 = - 27

x = \frac{17 + 129}{2}, x = \frac{17 - 129}{2}

Decimal Notation

x = 14.17890 \dots, x = 2.82109 \dots

Solution steps

x^{2} - 17 x + 13 = - 27

Move

27

to the left side

x^{2} - 17 x + 40 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 17 ) \pm ( - 17 ) ^{2} - 4 \cdot 1 \cdot 40}{2 \cdot 1}

(- 17)^{2} - 4 \cdot 1 \cdot 40 = 129

x_{1, 2} = \frac{- ( - 17 ) \pm 129}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 17 ) + 129}{2 \cdot 1}, x_{2} = \frac{- ( - 17 ) - 129}{2 \cdot 1}

x = \frac{- ( - 17 ) + 129}{2 \cdot 1} : \frac{17 + 129}{2}

x = \frac{- ( - 17 ) - 129}{2 \cdot 1} : \frac{17 - 129}{2}

The solutions to the quadratic equation are:

x = \frac{17 + 129}{2}, x = \frac{17 - 129}{2}

- \frac{1}{2} \cdot 8

(4 a + \frac{1}{8} b)^{2}

3 x + 2 \leq 14

\frac{1}{1. 1 ^{6}}

\frac{1}{a ^{2} + 2 a + 1} + \frac{1}{a + 1}

What are the solutions to the equation x^2-17x+13=-27 ?
The solutions to the equation x^2-17x+13=-27 are x=(17+sqrt(129))/2 ,x=(17-sqrt(129))/2
Find the zeros of x^2-17x+13=-27
The zeros of x^2-17x+13=-27 are x=(17+sqrt(129))/2 ,x=(17-sqrt(129))/2