What are the solutions to the equation 10x^2-47x+42=0 ?

The solutions to the equation 10x^2-47x+42=0 are x= 7/2 ,x= 6/5

Find the zeros of 10x^2-47x+42=0

The zeros of 10x^2-47x+42=0 are x= 7/2 ,x= 6/5

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10 x^{2} - 47 x + 42 = 0

x = \frac{7}{2}, x = \frac{6}{5}

Decimal Notation

x = 3.5, x = 1.2

Solution steps

10 x^{2} - 47 x + 42 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 47 ) \pm ( - 47 ) ^{2} - 4 \cdot 10 \cdot 42}{2 \cdot 10}

(- 47)^{2} - 4 \cdot 10 \cdot 42 = 23

x_{1, 2} = \frac{- ( - 47 ) \pm 23}{2 \cdot 10}

Separate the solutions

x_{1} = \frac{- ( - 47 ) + 23}{2 \cdot 10}, x_{2} = \frac{- ( - 47 ) - 23}{2 \cdot 10}

x = \frac{- ( - 47 ) + 23}{2 \cdot 10} : \frac{7}{2}

x = \frac{- ( - 47 ) - 23}{2 \cdot 10} : \frac{6}{5}

The solutions to the quadratic equation are:

x = \frac{7}{2}, x = \frac{6}{5}

x^{3} - 8 x

- 7 < - 1 - 6 x < 7

x^{3} - 5 x^{2} - 78 x - 72

2 > 8 - 43 h

x^{2} - x + \frac{1}{2} = 0

What are the solutions to the equation 10x^2-47x+42=0 ?
The solutions to the equation 10x^2-47x+42=0 are x= 7/2 ,x= 6/5
Find the zeros of 10x^2-47x+42=0
The zeros of 10x^2-47x+42=0 are x= 7/2 ,x= 6/5