What are the solutions to the equation (x^2-x+3)=5 ?

The solutions to the equation (x^2-x+3)=5 are x=2,x=-1

Find the zeros of (x^2-x+3)=5

The zeros of (x^2-x+3)=5 are x=2,x=-1

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(x^{2} - x + 3) = 5

x = 2, x = - 1

Solution steps

(x^{2} - x + 3) = 5

Refine

x^{2} - x + 3 = 5

Move

5

to the left side

x^{2} - x - 2 = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 2, x = - 1

x, f (x - 1) = x^{2} + 4 x - 5

x, z = 49 - 4 x^{2} - 4 y^{2}

(p + 4) (p + 4) = 25

15 x - 5 x^{2} = 0

y^{2} + y - 8 = 0

What are the solutions to the equation (x^2-x+3)=5 ?
The solutions to the equation (x^2-x+3)=5 are x=2,x=-1
Find the zeros of (x^2-x+3)=5
The zeros of (x^2-x+3)=5 are x=2,x=-1