What are the solutions to the equation 2x^2-28=x(x-3) ?

The solutions to the equation 2x^2-28=x(x-3) are x=4,x=-7

Find the zeros of 2x^2-28=x(x-3)

The zeros of 2x^2-28=x(x-3) are x=4,x=-7

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2 x^{2} - 28 = x (x - 3)

x = 4, x = - 7

Solution steps

2 x^{2} - 28 = x (x - 3)

Expand

x (x - 3) : x^{2} - 3 x

2 x^{2} - 28 = x^{2} - 3 x

Move

3 x

to the left side

2 x^{2} - 28 + 3 x = x^{2}

Move

x^{2}

to the left side

x^{2} + 3 x - 28 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot ( - 28 )}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot (- 28) = 11

x_{1, 2} = \frac{- 3 \pm 11}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 3 + 11}{2 \cdot 1}, x_{2} = \frac{- 3 - 11}{2 \cdot 1}

x = \frac{- 3 + 11}{2 \cdot 1} : 4

x = \frac{- 3 - 11}{2 \cdot 1} : - 7

The solutions to the quadratic equation are:

x = 4, x = - 7

2 x^{2} + 7 x - 3

a^{2} - 4 = 3 a

4 x^{2} + 27 x - 18 = 0

\frac{6 x ^{2} + 2 x}{3} = 20 x

x^{2} - 9 x = - 29

What are the solutions to the equation 2x^2-28=x(x-3) ?
The solutions to the equation 2x^2-28=x(x-3) are x=4,x=-7
Find the zeros of 2x^2-28=x(x-3)
The zeros of 2x^2-28=x(x-3) are x=4,x=-7