What are the solutions to the equation 42x^2-69x+20=7x^2-8 ?

The solutions to the equation 42x^2-69x+20=7x^2-8 are x= 7/5 ,x= 4/7

Find the zeros of 42x^2-69x+20=7x^2-8

The zeros of 42x^2-69x+20=7x^2-8 are x= 7/5 ,x= 4/7

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42 x^{2} - 69 x + 20 = 7 x^{2} - 8

x = \frac{7}{5}, x = \frac{4}{7}

Solution steps

42 x^{2} - 69 x + 20 = 7 x^{2} - 8

Move

8

to the left side

42 x^{2} - 69 x + 28 = 7 x^{2}

Move

7 x^{2}

to the left side

35 x^{2} - 69 x + 28 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 69 ) \pm ( - 69 ) ^{2} - 4 \cdot 35 \cdot 28}{2 \cdot 35}

(- 69)^{2} - 4 \cdot 35 \cdot 28 = 29

x_{1, 2} = \frac{- ( - 69 ) \pm 29}{2 \cdot 35}

Separate the solutions

x_{1} = \frac{- ( - 69 ) + 29}{2 \cdot 35}, x_{2} = \frac{- ( - 69 ) - 29}{2 \cdot 35}

x = \frac{- ( - 69 ) + 29}{2 \cdot 35} : \frac{7}{5}

x = \frac{- ( - 69 ) - 29}{2 \cdot 35} : \frac{4}{7}

The solutions to the quadratic equation are:

x = \frac{7}{5}, x = \frac{4}{7}

5 (6 - 3 Y) + 2 Y = 13

c^{2} - 100

- 3 ∣ x - 3 ∣ + 3 = 6

2 + \frac{9}{4}

(- \frac{5}{7})^{2} - \frac{5 ^{2}}{7}

What are the solutions to the equation 42x^2-69x+20=7x^2-8 ?
The solutions to the equation 42x^2-69x+20=7x^2-8 are x= 7/5 ,x= 4/7
Find the zeros of 42x^2-69x+20=7x^2-8
The zeros of 42x^2-69x+20=7x^2-8 are x= 7/5 ,x= 4/7