What are the solutions to the equation x^2-20x+12=0 ?

The solutions to the equation x^2-20x+12=0 are x=2(5+sqrt(22)),x=2(5-sqrt(22))

Find the zeros of x^2-20x+12=0

The zeros of x^2-20x+12=0 are x=2(5+sqrt(22)),x=2(5-sqrt(22))

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x^{2} - 20 x + 12 = 0

x = 2 (5 + 22), x = 2 (5 - 22)

Decimal Notation

x = 19.38083 \dots, x = 0.61916 \dots

Solution steps

x^{2} - 20 x + 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 \cdot 1 \cdot 12}{2 \cdot 1}

(- 20)^{2} - 4 \cdot 1 \cdot 12 = 422

x_{1, 2} = \frac{- ( - 20 ) \pm 4 22}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 20 ) + 4 22}{2 \cdot 1}, x_{2} = \frac{- ( - 20 ) - 4 22}{2 \cdot 1}

x = \frac{- ( - 20 ) + 4 22}{2 \cdot 1} : 2 (5 + 22)

x = \frac{- ( - 20 ) - 4 22}{2 \cdot 1} : 2 (5 - 22)

The solutions to the quadratic equation are:

x = 2 (5 + 22), x = 2 (5 - 22)

x, x^{2} + y^{2} = 0

r, V = π r^{2} h

3 x^{2} + 11 x + 7 = 0

2 x^{2} + 5 x + 20 = 0

x, z = x^{2} + y^{2}

What are the solutions to the equation x^2-20x+12=0 ?
The solutions to the equation x^2-20x+12=0 are x=2(5+sqrt(22)),x=2(5-sqrt(22))
Find the zeros of x^2-20x+12=0
The zeros of x^2-20x+12=0 are x=2(5+sqrt(22)),x=2(5-sqrt(22))