What are the solutions to the equation 5x^2-28x-4=8x-11 ?

The solutions to the equation 5x^2-28x-4=8x-11 are x=7,x= 1/5

Find the zeros of 5x^2-28x-4=8x-11

The zeros of 5x^2-28x-4=8x-11 are x=7,x= 1/5

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5 x^{2} - 28 x - 4 = 8 x - 11

x = 7, x = \frac{1}{5}

Decimal Notation

x = 7, x = 0.2

Solution steps

5 x^{2} - 28 x - 4 = 8 x - 11

Move

11

to the left side

5 x^{2} - 28 x + 7 = 8 x

Move

8 x

to the left side

5 x^{2} - 36 x + 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 36 ) \pm ( - 36 ) ^{2} - 4 \cdot 5 \cdot 7}{2 \cdot 5}

(- 36)^{2} - 4 \cdot 5 \cdot 7 = 34

x_{1, 2} = \frac{- ( - 36 ) \pm 34}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 36 ) + 34}{2 \cdot 5}, x_{2} = \frac{- ( - 36 ) - 34}{2 \cdot 5}

x = \frac{- ( - 36 ) + 34}{2 \cdot 5} : 7

x = \frac{- ( - 36 ) - 34}{2 \cdot 5} : \frac{1}{5}

The solutions to the quadratic equation are:

x = 7, x = \frac{1}{5}

5 p + 2 = - 10

4 x^{2} - 168 x + 36 = 0

r (r - 10) = 0

- 0.01 x^{2} + 0.04 x < x + 3

What are the solutions to the equation 5x^2-28x-4=8x-11 ?
The solutions to the equation 5x^2-28x-4=8x-11 are x=7,x= 1/5
Find the zeros of 5x^2-28x-4=8x-11
The zeros of 5x^2-28x-4=8x-11 are x=7,x= 1/5