What are the solutions to the equation 1x^2-11x+15=0 ?

The solutions to the equation 1x^2-11x+15=0 are x=(11+sqrt(61))/2 ,x=(11-sqrt(61))/2

Find the zeros of 1x^2-11x+15=0

The zeros of 1x^2-11x+15=0 are x=(11+sqrt(61))/2 ,x=(11-sqrt(61))/2

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1 x^{2} - 11 x + 15 = 0

x = \frac{11 + 61}{2}, x = \frac{11 - 61}{2}

Decimal Notation

x = 9.40512 \dots, x = 1.59487 \dots

Solution steps

1 \cdot x^{2} - 11 x + 15 = 0

Refine

x^{2} - 11 x + 15 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 1 \cdot 15}{2 \cdot 1}

(- 11)^{2} - 4 \cdot 1 \cdot 15 = 61

x_{1, 2} = \frac{- ( - 11 ) \pm 61}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 11 ) + 61}{2 \cdot 1}, x_{2} = \frac{- ( - 11 ) - 61}{2 \cdot 1}

x = \frac{- ( - 11 ) + 61}{2 \cdot 1} : \frac{11 + 61}{2}

x = \frac{- ( - 11 ) - 61}{2 \cdot 1} : \frac{11 - 61}{2}

The solutions to the quadratic equation are:

x = \frac{11 + 61}{2}, x = \frac{11 - 61}{2}

416

36 x^{2} - 96 x + 64

\frac{x - 8}{5} = \frac{x + 4}{9}

2 < \frac{1}{3} - 2 x < 5

What are the solutions to the equation 1x^2-11x+15=0 ?
The solutions to the equation 1x^2-11x+15=0 are x=(11+sqrt(61))/2 ,x=(11-sqrt(61))/2
Find the zeros of 1x^2-11x+15=0
The zeros of 1x^2-11x+15=0 are x=(11+sqrt(61))/2 ,x=(11-sqrt(61))/2