What are the solutions to the equation 7x^2-27x+1=5 ?

The solutions to the equation 7x^2-27x+1=5 are x=4,x=-1/7

Find the zeros of 7x^2-27x+1=5

The zeros of 7x^2-27x+1=5 are x=4,x=-1/7

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7 x^{2} - 27 x + 1 = 5

x = 4, x = - \frac{1}{7}

Decimal Notation

x = 4, x = - 0.14285 \dots

Solution steps

7 x^{2} - 27 x + 1 = 5

Move

5

to the left side

7 x^{2} - 27 x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 27 ) \pm ( - 27 ) ^{2} - 4 \cdot 7 ( - 4 )}{2 \cdot 7}

(- 27)^{2} - 4 \cdot 7 (- 4) = 29

x_{1, 2} = \frac{- ( - 27 ) \pm 29}{2 \cdot 7}

Separate the solutions

x_{1} = \frac{- ( - 27 ) + 29}{2 \cdot 7}, x_{2} = \frac{- ( - 27 ) - 29}{2 \cdot 7}

x = \frac{- ( - 27 ) + 29}{2 \cdot 7} : 4

x = \frac{- ( - 27 ) - 29}{2 \cdot 7} : - \frac{1}{7}

The solutions to the quadratic equation are:

x = 4, x = - \frac{1}{7}

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

x^{2} + 10 x + 14 = 0

x^{2} + 4 x = 0

(x^{2} - 12 x + 35) = 0

x^{2} + 10 x - 2000 = 0

What are the solutions to the equation 7x^2-27x+1=5 ?
The solutions to the equation 7x^2-27x+1=5 are x=4,x=-1/7
Find the zeros of 7x^2-27x+1=5
The zeros of 7x^2-27x+1=5 are x=4,x=-1/7