What are the solutions to the equation 12=-3x^2+16x ?

The solutions to the equation 12=-3x^2+16x are x=(2(4-sqrt(7)))/3 ,x=(2(4+sqrt(7)))/3

Find the zeros of 12=-3x^2+16x

The zeros of 12=-3x^2+16x are x=(2(4-sqrt(7)))/3 ,x=(2(4+sqrt(7)))/3

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12 = - 3 x^{2} + 16 x

x = \frac{2 ( 4 - 7 )}{3}, x = \frac{2 ( 4 + 7 )}{3}

Decimal Notation

x = 0.90283 \dots, x = 4.43050 \dots

Solution steps

12 = - 3 x^{2} + 16 x

Switch sides

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

Move

12

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 ( - 3 ) ( - 12 )}{2 ( - 3 )}

1 6^{2} - 4 (- 3) (- 12) = 47

x_{1, 2} = \frac{- 16 \pm 4 7}{2 ( - 3 )}

Separate the solutions

x_{1} = \frac{- 16 + 4 7}{2 ( - 3 )}, x_{2} = \frac{- 16 - 4 7}{2 ( - 3 )}

x = \frac{- 16 + 4 7}{2 ( - 3 )} : \frac{2 ( 4 - 7 )}{3}

x = \frac{- 16 - 4 7}{2 ( - 3 )} : \frac{2 ( 4 + 7 )}{3}

The solutions to the quadratic equation are:

x = \frac{2 ( 4 - 7 )}{3}, x = \frac{2 ( 4 + 7 )}{3}

x, \frac{x ^{2}}{25} + y^{2} = - 1

x^{2} + 2 x - 11 = - 6

3 (7 x - 2) + 3 x (x - 4) = 1

x^{2} = 1^{2} + 2^{2}

(k + 7)^{2} = 289

What are the solutions to the equation 12=-3x^2+16x ?
The solutions to the equation 12=-3x^2+16x are x=(2(4-sqrt(7)))/3 ,x=(2(4+sqrt(7)))/3
Find the zeros of 12=-3x^2+16x
The zeros of 12=-3x^2+16x are x=(2(4-sqrt(7)))/3 ,x=(2(4+sqrt(7)))/3