What are the solutions to the equation x^2-16x+12=0 ?

The solutions to the equation x^2-16x+12=0 are x=2(4+sqrt(13)),x=2(4-sqrt(13))

Find the zeros of x^2-16x+12=0

The zeros of x^2-16x+12=0 are x=2(4+sqrt(13)),x=2(4-sqrt(13))

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x^{2} - 16 x + 12 = 0

x = 2 (4 + 13), x = 2 (4 - 13)

Decimal Notation

x = 15.21110 \dots, x = 0.78889 \dots

Solution steps

x^{2} - 16 x + 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 1 \cdot 12}{2 \cdot 1}

(- 16)^{2} - 4 \cdot 1 \cdot 12 = 413

x_{1, 2} = \frac{- ( - 16 ) \pm 4 13}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 16 ) + 4 13}{2 \cdot 1}, x_{2} = \frac{- ( - 16 ) - 4 13}{2 \cdot 1}

x = \frac{- ( - 16 ) + 4 13}{2 \cdot 1} : 2 (4 + 13)

x = \frac{- ( - 16 ) - 4 13}{2 \cdot 1} : 2 (4 - 13)

The solutions to the quadratic equation are:

x = 2 (4 + 13), x = 2 (4 - 13)

x^{2} + 3 x - 10 = 0

- 4 x^{2} - 12 x + 5 = 0

s^{2} + 6 s + 12 = 0

- 5 x^{2} + 95 x + 100 = 0

a x^{2} = 4 x - x^{2}

What are the solutions to the equation x^2-16x+12=0 ?
The solutions to the equation x^2-16x+12=0 are x=2(4+sqrt(13)),x=2(4-sqrt(13))
Find the zeros of x^2-16x+12=0
The zeros of x^2-16x+12=0 are x=2(4+sqrt(13)),x=2(4-sqrt(13))