What are the solutions to the equation 2x^2-20x+42=0 ?

The solutions to the equation 2x^2-20x+42=0 are x=7,x=3

Find the zeros of 2x^2-20x+42=0

The zeros of 2x^2-20x+42=0 are x=7,x=3

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2 x^{2} - 20 x + 42 = 0

x = 7, x = 3

Solution steps

2 x^{2} - 20 x + 42 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 \cdot 2 \cdot 42}{2 \cdot 2}

(- 20)^{2} - 4 \cdot 2 \cdot 42 = 8

x_{1, 2} = \frac{- ( - 20 ) \pm 8}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 20 ) + 8}{2 \cdot 2}, x_{2} = \frac{- ( - 20 ) - 8}{2 \cdot 2}

x = \frac{- ( - 20 ) + 8}{2 \cdot 2} : 7

x = \frac{- ( - 20 ) - 8}{2 \cdot 2} : 3

The solutions to the quadratic equation are:

x = 7, x = 3

(- 1)^{2} + 2 (- 1) + 1

4 p + 2 < 2 (p + 5)

3 x - 1 \leq - 10

3 x^{2} - 10 x + 15 = 0

2.7118200000000 \dots x^{2} - 0.7182818285 x - 0.1323621525 = 0

What are the solutions to the equation 2x^2-20x+42=0 ?
The solutions to the equation 2x^2-20x+42=0 are x=7,x=3
Find the zeros of 2x^2-20x+42=0
The zeros of 2x^2-20x+42=0 are x=7,x=3