What are the solutions to the equation 26=x^2-12x ?

The solutions to the equation 26=x^2-12x are x=6+sqrt(62),x=6-sqrt(62)

Find the zeros of 26=x^2-12x

The zeros of 26=x^2-12x are x=6+sqrt(62),x=6-sqrt(62)

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26 = x^{2} - 12 x

x = 6 + 62, x = 6 - 62

Decimal Notation

x = 13.87400 \dots, x = - 1.87400 \dots

Solution steps

26 = x^{2} - 12 x

Switch sides

x^{2} - 12 x = 26

Move

26

to the left side

x^{2} - 12 x - 26 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 1 \cdot ( - 26 )}{2 \cdot 1}

(- 12)^{2} - 4 \cdot 1 \cdot (- 26) = 262

x_{1, 2} = \frac{- ( - 12 ) \pm 2 62}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 2 62}{2 \cdot 1}, x_{2} = \frac{- ( - 12 ) - 2 62}{2 \cdot 1}

x = \frac{- ( - 12 ) + 2 62}{2 \cdot 1} : 6 + 62

x = \frac{- ( - 12 ) - 2 62}{2 \cdot 1} : 6 - 62

The solutions to the quadratic equation are:

x = 6 + 62, x = 6 - 62

\frac{1}{7 ^{2}}

\frac{7}{7 ^{\frac{1}{3}}}

(x + 3)^{2} = (x - 1)^{2}

- 5 p^{2} - 40 p = 64

x - 4 < - 3

What are the solutions to the equation 26=x^2-12x ?
The solutions to the equation 26=x^2-12x are x=6+sqrt(62),x=6-sqrt(62)
Find the zeros of 26=x^2-12x
The zeros of 26=x^2-12x are x=6+sqrt(62),x=6-sqrt(62)