What are the solutions to the equation 2x^2-11x+28=x^2 ?

The solutions to the equation 2x^2-11x+28=x^2 are x=7,x=4

Find the zeros of 2x^2-11x+28=x^2

The zeros of 2x^2-11x+28=x^2 are x=7,x=4

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2 x^{2} - 11 x + 28 = x^{2}

x = 7, x = 4

Solution steps

2 x^{2} - 11 x + 28 = x^{2}

Move

x^{2}

to the left side

x^{2} - 11 x + 28 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 1 \cdot 28}{2 \cdot 1}

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 7, x = 4

y, 9 x^{2} + 4 y^{2} = 36

6 x^{2} - 35 = - 11 x

2 x^{2} + 7 x - 5 = 0

x, x^{2} - y - 1 = C

x, 4 x^{2} + 8 x = 21

What are the solutions to the equation 2x^2-11x+28=x^2 ?
The solutions to the equation 2x^2-11x+28=x^2 are x=7,x=4
Find the zeros of 2x^2-11x+28=x^2
The zeros of 2x^2-11x+28=x^2 are x=7,x=4