What are the solutions to the equation 52-x^2+x=10 ?

The solutions to the equation 52-x^2+x=10 are x=-6,x=7

Find the zeros of 52-x^2+x=10

The zeros of 52-x^2+x=10 are x=-6,x=7

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52 - x^{2} + x = 10

x = - 6, x = 7

Solution steps

52 - x^{2} + x = 10

Move

10

to the left side

- x^{2} + x + 42 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 1 ) \cdot 42}{2 ( - 1 )}

1^{2} - 4 (- 1) \cdot 42 = 13

x_{1, 2} = \frac{- 1 \pm 13}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 1 + 13}{2 ( - 1 )}, x_{2} = \frac{- 1 - 13}{2 ( - 1 )}

x = \frac{- 1 + 13}{2 ( - 1 )} : - 6

x = \frac{- 1 - 13}{2 ( - 1 )} : 7

The solutions to the quadratic equation are:

x = - 6, x = 7

x^{3} - 9 x^{2} + 23 x - 15

1000 0^{\frac{1}{4}}

5 x > 35

14 b = - 56

\frac{3.9}{3.9 - 4}

What are the solutions to the equation 52-x^2+x=10 ?
The solutions to the equation 52-x^2+x=10 are x=-6,x=7
Find the zeros of 52-x^2+x=10
The zeros of 52-x^2+x=10 are x=-6,x=7