What are the solutions to the equation x+2x^2-4=-7x+6 ?

The solutions to the equation x+2x^2-4=-7x+6 are x=1,x=-5

Find the zeros of x+2x^2-4=-7x+6

The zeros of x+2x^2-4=-7x+6 are x=1,x=-5

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x + 2 x^{2} - 4 = - 7 x + 6

x = 1, x = - 5

Solution steps

x + 2 x^{2} - 4 = - 7 x + 6

Move

6

to the left side

2 x^{2} + x - 10 = - 7 x

Move

7 x

to the left side

2 x^{2} + 8 x - 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 2 ( - 10 )}{2 \cdot 2}

8^{2} - 4 \cdot 2 (- 10) = 12

x_{1, 2} = \frac{- 8 \pm 12}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 8 + 12}{2 \cdot 2}, x_{2} = \frac{- 8 - 12}{2 \cdot 2}

x = \frac{- 8 + 12}{2 \cdot 2} : 1

x = \frac{- 8 - 12}{2 \cdot 2} : - 5

The solutions to the quadratic equation are:

x = 1, x = - 5

(x - 1) (x + 3) = 6

x + 2 = \frac{( x - 2 ) ^{2}}{2}

x, k x^{2} - 3 k x + 9 = 0

5 y^{2} - 2 = - 9 y

2 (3 x - 1)^{2} = 162

What are the solutions to the equation x+2x^2-4=-7x+6 ?
The solutions to the equation x+2x^2-4=-7x+6 are x=1,x=-5
Find the zeros of x+2x^2-4=-7x+6
The zeros of x+2x^2-4=-7x+6 are x=1,x=-5