What are the solutions to the equation 3x^2-16x+13=-4 ?

The solutions to the equation 3x^2-16x+13=-4 are x=(8+sqrt(13))/3 ,x=(8-sqrt(13))/3

Find the zeros of 3x^2-16x+13=-4

The zeros of 3x^2-16x+13=-4 are x=(8+sqrt(13))/3 ,x=(8-sqrt(13))/3

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3 x^{2} - 16 x + 13 = - 4

x = \frac{8 + 13}{3}, x = \frac{8 - 13}{3}

Solution steps

3 x^{2} - 16 x + 13 = - 4

Move

4

to the left side

3 x^{2} - 16 x + 17 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 3 \cdot 17}{2 \cdot 3}

(- 16)^{2} - 4 \cdot 3 \cdot 17 = 213

x_{1, 2} = \frac{- ( - 16 ) \pm 2 13}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 16 ) + 2 13}{2 \cdot 3}, x_{2} = \frac{- ( - 16 ) - 2 13}{2 \cdot 3}

x = \frac{- ( - 16 ) + 2 13}{2 \cdot 3} : \frac{8 + 13}{3}

x = \frac{- ( - 16 ) - 2 13}{2 \cdot 3} : \frac{8 - 13}{3}

The solutions to the quadratic equation are:

x = \frac{8 + 13}{3}, x = \frac{8 - 13}{3}

1 3^{2} + 1 2^{2} = x^{2}

0 = x^{2} - 13 x + 36

p^{2} - 5 p + 6 = 0

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

x, 3 x^{2} + y = 3 y^{2} + x

What are the solutions to the equation 3x^2-16x+13=-4 ?
The solutions to the equation 3x^2-16x+13=-4 are x=(8+sqrt(13))/3 ,x=(8-sqrt(13))/3
Find the zeros of 3x^2-16x+13=-4
The zeros of 3x^2-16x+13=-4 are x=(8+sqrt(13))/3 ,x=(8-sqrt(13))/3