What are the solutions to the equation 56=x^2-x ?

The solutions to the equation 56=x^2-x are x=8,x=-7

Find the zeros of 56=x^2-x

The zeros of 56=x^2-x are x=8,x=-7

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56 = x^{2} - x

x = 8, x = - 7

Solution steps

56 = x^{2} - x

Switch sides

x^{2} - x = 56

Move

56

to the left side

x^{2} - x - 56 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 56 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 56) = 15

x_{1, 2} = \frac{- ( - 1 ) \pm 15}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 15}{2 \cdot 1}, x_{2} = \frac{- ( - 1 ) - 15}{2 \cdot 1}

x = \frac{- ( - 1 ) + 15}{2 \cdot 1} : 8

x = \frac{- ( - 1 ) - 15}{2 \cdot 1} : - 7

The solutions to the quadratic equation are:

x = 8, x = - 7

2 (x - 8) - 6 x = - 4 - 4 (4 x - 6)

8 = x + 5

7 + 5 y = 22

40 \leq \frac{5}{9} (F - 32) \leq 50

n^{2} - 3 n - 40

What are the solutions to the equation 56=x^2-x ?
The solutions to the equation 56=x^2-x are x=8,x=-7
Find the zeros of 56=x^2-x
The zeros of 56=x^2-x are x=8,x=-7