What are the solutions to the equation x^2-20x+28=0 ?

The solutions to the equation x^2-20x+28=0 are x=2(5+3sqrt(2)),x=2(5-3sqrt(2))

Find the zeros of x^2-20x+28=0

The zeros of x^2-20x+28=0 are x=2(5+3sqrt(2)),x=2(5-3sqrt(2))

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x^{2} - 20 x + 28 = 0

x = 2 (5 + 32), x = 2 (5 - 32)

Decimal Notation

x = 18.48528 \dots, x = 1.51471 \dots

Solution steps

x^{2} - 20 x + 28 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 \cdot 1 \cdot 28}{2 \cdot 1}

(- 20)^{2} - 4 \cdot 1 \cdot 28 = 122

x_{1, 2} = \frac{- ( - 20 ) \pm 12 2}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 20 ) + 12 2}{2 \cdot 1}, x_{2} = \frac{- ( - 20 ) - 12 2}{2 \cdot 1}

x = \frac{- ( - 20 ) + 12 2}{2 \cdot 1} : 2 (5 + 32)

x = \frac{- ( - 20 ) - 12 2}{2 \cdot 1} : 2 (5 - 32)

The solutions to the quadratic equation are:

x = 2 (5 + 32), x = 2 (5 - 32)

5^{2} = 3^{2} + x^{2}

x^{2} + 8 x - 20 = 0

x^{2} + 3 x = 0

6 x^{2} - 24 x + 18 = 0

2 x^{2} + 8 x - 3

What are the solutions to the equation x^2-20x+28=0 ?
The solutions to the equation x^2-20x+28=0 are x=2(5+3sqrt(2)),x=2(5-3sqrt(2))
Find the zeros of x^2-20x+28=0
The zeros of x^2-20x+28=0 are x=2(5+3sqrt(2)),x=2(5-3sqrt(2))