What are the solutions to the equation 12x=13-x^2 ?

The solutions to the equation 12x=13-x^2 are x=-13,x=1

Find the zeros of 12x=13-x^2

The zeros of 12x=13-x^2 are x=-13,x=1

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12 x = 13 - x^{2}

x = - 13, x = 1

Solution steps

12 x = 13 - x^{2}

Switch sides

13 - x^{2} = 12 x

Move

12 x

to the left side

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

Write in the standard form

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

- x^{2} - 12 x + 13 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 ( - 1 ) \cdot 13}{2 ( - 1 )}

(- 12)^{2} - 4 (- 1) \cdot 13 = 14

x_{1, 2} = \frac{- ( - 12 ) \pm 14}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 14}{2 ( - 1 )}, x_{2} = \frac{- ( - 12 ) - 14}{2 ( - 1 )}

x = \frac{- ( - 12 ) + 14}{2 ( - 1 )} : - 13

x = \frac{- ( - 12 ) - 14}{2 ( - 1 )} : 1

The solutions to the quadratic equation are:

x = - 13, x = 1

- 2 x^{2} + 7 x + 1 = 17

t^{2} - t + 1 = 0

x^{2} + 3 x - 2

x (x + 1) = 1

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

What are the solutions to the equation 12x=13-x^2 ?
The solutions to the equation 12x=13-x^2 are x=-13,x=1
Find the zeros of 12x=13-x^2
The zeros of 12x=13-x^2 are x=-13,x=1