What are the solutions to the equation 5x^2-4x-7=1-x ?

The solutions to the equation 5x^2-4x-7=1-x are x= 8/5 ,x=-1

Find the zeros of 5x^2-4x-7=1-x

The zeros of 5x^2-4x-7=1-x are x= 8/5 ,x=-1

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5 x^{2} - 4 x - 7 = 1 - x

x = \frac{8}{5}, x = - 1

Decimal Notation

x = 1.6, x = - 1

Solution steps

5 x^{2} - 4 x - 7 = 1 - x

Move

x

to the left side

5 x^{2} - 3 x - 7 = 1

Move

1

to the left side

5 x^{2} - 3 x - 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 5 ( - 8 )}{2 \cdot 5}

(- 3)^{2} - 4 \cdot 5 (- 8) = 13

x_{1, 2} = \frac{- ( - 3 ) \pm 13}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 13}{2 \cdot 5}, x_{2} = \frac{- ( - 3 ) - 13}{2 \cdot 5}

x = \frac{- ( - 3 ) + 13}{2 \cdot 5} : \frac{8}{5}

x = \frac{- ( - 3 ) - 13}{2 \cdot 5} : - 1

The solutions to the quadratic equation are:

x = \frac{8}{5}, x = - 1

x, x^{2} - x - 2 = 0

u^{2} - 17 u + 16 = 0

16 x^{2} + 9 = 24 x

- 2 c^{2} + 45 = 27

6 x^{2} + 120 x + 2070 = 0

What are the solutions to the equation 5x^2-4x-7=1-x ?
The solutions to the equation 5x^2-4x-7=1-x are x= 8/5 ,x=-1
Find the zeros of 5x^2-4x-7=1-x
The zeros of 5x^2-4x-7=1-x are x= 8/5 ,x=-1