What are the solutions to the equation x^2-36=12x ?

The solutions to the equation x^2-36=12x are x=6(1+sqrt(2)),x=6(1-sqrt(2))

Find the zeros of x^2-36=12x

The zeros of x^2-36=12x are x=6(1+sqrt(2)),x=6(1-sqrt(2))

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x^{2} - 36 = 12 x

x = 6 (1 + 2), x = 6 (1 - 2)

Decimal Notation

x = 14.48528 \dots, x = - 2.48528 \dots

Solution steps

x^{2} - 36 = 12 x

Move

12 x

to the left side

x^{2} - 36 - 12 x = 0

Write in the standard form

a x^{2} + b x + c = 0

x^{2} - 12 x - 36 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 1 \cdot ( - 36 )}{2 \cdot 1}

(- 12)^{2} - 4 \cdot 1 \cdot (- 36) = 122

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

Separate the solutions

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

x = \frac{- ( - 12 ) + 12 2}{2 \cdot 1} : 6 (1 + 2)

x = \frac{- ( - 12 ) - 12 2}{2 \cdot 1} : 6 (1 - 2)

The solutions to the quadratic equation are:

x = 6 (1 + 2), x = 6 (1 - 2)

4 + 2 (5 x + 2 x^{2}) = 11

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

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

x^{2} - 5 x = 8

y, x^{2} + y^{2} = \frac{2 y}{x}

What are the solutions to the equation x^2-36=12x ?
The solutions to the equation x^2-36=12x are x=6(1+sqrt(2)),x=6(1-sqrt(2))
Find the zeros of x^2-36=12x
The zeros of x^2-36=12x are x=6(1+sqrt(2)),x=6(1-sqrt(2))