What are the solutions to the equation -13x^2-9x+4=5+2x^2-x ?

The solutions to the equation -13x^2-9x+4=5+2x^2-x are x=-1/3 ,x=-1/5

Find the zeros of -13x^2-9x+4=5+2x^2-x

The zeros of -13x^2-9x+4=5+2x^2-x are x=-1/3 ,x=-1/5

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- 13 x^{2} - 9 x + 4 = 5 + 2 x^{2} - x

x = - \frac{1}{3}, x = - \frac{1}{5}

Decimal Notation

x = - 0.33333 \dots, x = - 0.2

Solution steps

- 13 x^{2} - 9 x + 4 = 5 + 2 x^{2} - x

Move

x

to the left side

- 13 x^{2} - 8 x + 4 = 5 + 2 x^{2}

Move

2 x^{2}

to the left side

- 15 x^{2} - 8 x + 4 = 5

Move

5

to the left side

- 15 x^{2} - 8 x - 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 ( - 15 ) ( - 1 )}{2 ( - 15 )}

(- 8)^{2} - 4 (- 15) (- 1) = 2

x_{1, 2} = \frac{- ( - 8 ) \pm 2}{2 ( - 15 )}

Separate the solutions

x_{1} = \frac{- ( - 8 ) + 2}{2 ( - 15 )}, x_{2} = \frac{- ( - 8 ) - 2}{2 ( - 15 )}

x = \frac{- ( - 8 ) + 2}{2 ( - 15 )} : - \frac{1}{3}

x = \frac{- ( - 8 ) - 2}{2 ( - 15 )} : - \frac{1}{5}

The solutions to the quadratic equation are:

x = - \frac{1}{3}, x = - \frac{1}{5}

4 x^{2} - 25 \leq 0

(- 11)^{4}

2 x^{3} - 4 x^{2}

a^{2} + ab

What are the solutions to the equation -13x^2-9x+4=5+2x^2-x ?
The solutions to the equation -13x^2-9x+4=5+2x^2-x are x=-1/3 ,x=-1/5
Find the zeros of -13x^2-9x+4=5+2x^2-x
The zeros of -13x^2-9x+4=5+2x^2-x are x=-1/3 ,x=-1/5