What are the solutions to the equation x^2-7x+8=0 ?

The solutions to the equation x^2-7x+8=0 are x=(7+sqrt(17))/2 ,x=(7-sqrt(17))/2

Find the zeros of x^2-7x+8=0

The zeros of x^2-7x+8=0 are x=(7+sqrt(17))/2 ,x=(7-sqrt(17))/2

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x^{2} - 7 x + 8 = 0

x = \frac{7 + 17}{2}, x = \frac{7 - 17}{2}

Decimal Notation

x = 5.56155 \dots, x = 1.43844 \dots

Solution steps

x^{2} - 7 x + 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 1 \cdot 8}{2 \cdot 1}

(- 7)^{2} - 4 \cdot 1 \cdot 8 = 17

x_{1, 2} = \frac{- ( - 7 ) \pm 17}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 17}{2 \cdot 1}, x_{2} = \frac{- ( - 7 ) - 17}{2 \cdot 1}

x = \frac{- ( - 7 ) + 17}{2 \cdot 1} : \frac{7 + 17}{2}

x = \frac{- ( - 7 ) - 17}{2 \cdot 1} : \frac{7 - 17}{2}

The solutions to the quadratic equation are:

x = \frac{7 + 17}{2}, x = \frac{7 - 17}{2}

2 - \frac{2 x + 3}{2} = \frac{x + 6}{8} - x

\frac{3}{2 - 3}

\frac{1}{3 ^{x} + 5} < \frac{1}{3 ^{x + 1} - 1}

(2 t^{- 3})^{3} (0.4 r)^{2}

\frac{x ^{2} - 9}{x - 1} \leq 0

What are the solutions to the equation x^2-7x+8=0 ?
The solutions to the equation x^2-7x+8=0 are x=(7+sqrt(17))/2 ,x=(7-sqrt(17))/2
Find the zeros of x^2-7x+8=0
The zeros of x^2-7x+8=0 are x=(7+sqrt(17))/2 ,x=(7-sqrt(17))/2