What are the solutions to the equation x^2-14x+31=63 ?

The solutions to the equation x^2-14x+31=63 are x=16,x=-2

Find the zeros of x^2-14x+31=63

The zeros of x^2-14x+31=63 are x=16,x=-2

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x^{2} - 14 x + 31 = 63

x = 16, x = - 2

Solution steps

x^{2} - 14 x + 31 = 63

Move

63

to the left side

x^{2} - 14 x - 32 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 14 ) \pm ( - 14 ) ^{2} - 4 \cdot 1 \cdot ( - 32 )}{2 \cdot 1}

(- 14)^{2} - 4 \cdot 1 \cdot (- 32) = 18

x_{1, 2} = \frac{- ( - 14 ) \pm 18}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 14 ) + 18}{2 \cdot 1}, x_{2} = \frac{- ( - 14 ) - 18}{2 \cdot 1}

x = \frac{- ( - 14 ) + 18}{2 \cdot 1} : 16

x = \frac{- ( - 14 ) - 18}{2 \cdot 1} : - 2

The solutions to the quadratic equation are:

x = 16, x = - 2

x^{2} - 18 x - 19

x^{2} - 4 x = 45

2 (1 - x) + 7 \leq 4 (2 x - 1)

(x - 7) (4 - x) (x + 1) < 0

9 h - 2 \leq 10 h - 5

What are the solutions to the equation x^2-14x+31=63 ?
The solutions to the equation x^2-14x+31=63 are x=16,x=-2
Find the zeros of x^2-14x+31=63
The zeros of x^2-14x+31=63 are x=16,x=-2