What are the solutions to the equation 2x^2-15x+13=1-x^2 ?

The solutions to the equation 2x^2-15x+13=1-x^2 are x=4,x=1

Find the zeros of 2x^2-15x+13=1-x^2

The zeros of 2x^2-15x+13=1-x^2 are x=4,x=1

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2 x^{2} - 15 x + 13 = 1 - x^{2}

x = 4, x = 1

Solution steps

2 x^{2} - 15 x + 13 = 1 - x^{2}

Move

x^{2}

to the left side

3 x^{2} - 15 x + 13 = 1

Move

1

to the left side

3 x^{2} - 15 x + 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 15 ) \pm ( - 15 ) ^{2} - 4 \cdot 3 \cdot 12}{2 \cdot 3}

(- 15)^{2} - 4 \cdot 3 \cdot 12 = 9

x_{1, 2} = \frac{- ( - 15 ) \pm 9}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 15 ) + 9}{2 \cdot 3}, x_{2} = \frac{- ( - 15 ) - 9}{2 \cdot 3}

x = \frac{- ( - 15 ) + 9}{2 \cdot 3} : 4

x = \frac{- ( - 15 ) - 9}{2 \cdot 3} : 1

The solutions to the quadratic equation are:

x = 4, x = 1

3^{2} + b^{2} = 7^{2}

2 x^{2} - 5 x + 67 = 0

3 x^{2} + 6 x = 8

x^{2} - \frac{1}{6} x - \frac{1}{6} = 0

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

What are the solutions to the equation 2x^2-15x+13=1-x^2 ?
The solutions to the equation 2x^2-15x+13=1-x^2 are x=4,x=1
Find the zeros of 2x^2-15x+13=1-x^2
The zeros of 2x^2-15x+13=1-x^2 are x=4,x=1