What are the solutions to the equation 2x^2=-6+8x ?

The solutions to the equation 2x^2=-6+8x are x=3,x=1

Find the zeros of 2x^2=-6+8x

The zeros of 2x^2=-6+8x are x=3,x=1

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2 x^{2} = - 6 + 8 x

x = 3, x = 1

Solution steps

2 x^{2} = - 6 + 8 x

Move

8 x

to the left side

2 x^{2} - 8 x = - 6

Move

6

to the left side

2 x^{2} - 8 x + 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 2 \cdot 6}{2 \cdot 2}

(- 8)^{2} - 4 \cdot 2 \cdot 6 = 4

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

Separate the solutions

x_{1} = \frac{- ( - 8 ) + 4}{2 \cdot 2}, x_{2} = \frac{- ( - 8 ) - 4}{2 \cdot 2}

x = \frac{- ( - 8 ) + 4}{2 \cdot 2} : 3

x = \frac{- ( - 8 ) - 4}{2 \cdot 2} : 1

The solutions to the quadratic equation are:

x = 3, x = 1

- 5 > r - 630

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

(3 x + 2) (3 x + 5) - (3 x + 9) (3 x - 2) = x

6 t^{2} + 13 t + 6

\frac{π}{4} - \frac{π}{16}

What are the solutions to the equation 2x^2=-6+8x ?
The solutions to the equation 2x^2=-6+8x are x=3,x=1
Find the zeros of 2x^2=-6+8x
The zeros of 2x^2=-6+8x are x=3,x=1