What are the solutions to the equation 35x^2-22x+7=4 ?

The solutions to the equation 35x^2-22x+7=4 are x= 3/7 ,x= 1/5

Find the zeros of 35x^2-22x+7=4

The zeros of 35x^2-22x+7=4 are x= 3/7 ,x= 1/5

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35 x^{2} - 22 x + 7 = 4

x = \frac{3}{7}, x = \frac{1}{5}

Decimal Notation

x = 0.42857 \dots, x = 0.2

Solution steps

35 x^{2} - 22 x + 7 = 4

Move

4

to the left side

35 x^{2} - 22 x + 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 22 ) \pm ( - 22 ) ^{2} - 4 \cdot 35 \cdot 3}{2 \cdot 35}

(- 22)^{2} - 4 \cdot 35 \cdot 3 = 8

x_{1, 2} = \frac{- ( - 22 ) \pm 8}{2 \cdot 35}

Separate the solutions

x_{1} = \frac{- ( - 22 ) + 8}{2 \cdot 35}, x_{2} = \frac{- ( - 22 ) - 8}{2 \cdot 35}

x = \frac{- ( - 22 ) + 8}{2 \cdot 35} : \frac{3}{7}

x = \frac{- ( - 22 ) - 8}{2 \cdot 35} : \frac{1}{5}

The solutions to the quadratic equation are:

x = \frac{3}{7}, x = \frac{1}{5}

∣ x - 4 ∣ + 1 < 5

- 14 (x - 5) < - 84

3 x^{2} + 6 x - 75 = 0

10 k^{4} + 7 k^{3} - 3 k^{2}

20 x^{2} + 3 x + 1 = 0

What are the solutions to the equation 35x^2-22x+7=4 ?
The solutions to the equation 35x^2-22x+7=4 are x= 3/7 ,x= 1/5
Find the zeros of 35x^2-22x+7=4
The zeros of 35x^2-22x+7=4 are x= 3/7 ,x= 1/5