What are the solutions to the equation 5x^2-3x+1=10 ?

The solutions to the equation 5x^2-3x+1=10 are x=(3+3sqrt(21))/(10),x=(3-3sqrt(21))/(10)

Find the zeros of 5x^2-3x+1=10

The zeros of 5x^2-3x+1=10 are x=(3+3sqrt(21))/(10),x=(3-3sqrt(21))/(10)

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5 x^{2} - 3 x + 1 = 10

x = \frac{3 + 3 21}{10}, x = \frac{3 - 3 21}{10}

Decimal Notation

x = 1.67477 \dots, x = - 1.07477 \dots

Solution steps

5 x^{2} - 3 x + 1 = 10

Move

10

to the left side

5 x^{2} - 3 x - 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 5 ( - 9 )}{2 \cdot 5}

(- 3)^{2} - 4 \cdot 5 (- 9) = 321

x_{1, 2} = \frac{- ( - 3 ) \pm 3 21}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 3 21}{2 \cdot 5}, x_{2} = \frac{- ( - 3 ) - 3 21}{2 \cdot 5}

x = \frac{- ( - 3 ) + 3 21}{2 \cdot 5} : \frac{3 + 3 21}{10}

x = \frac{- ( - 3 ) - 3 21}{2 \cdot 5} : \frac{3 - 3 21}{10}

The solutions to the quadratic equation are:

x = \frac{3 + 3 21}{10}, x = \frac{3 - 3 21}{10}

7 x^{2} = - 9 x - 3

m^{2} + 6 m = - 8

x^{2} + 10 x - 3

- 6 = x^{2} + 4 x - 1

x, y^{2} + 2 x y + 6 x^{2} = 8

What are the solutions to the equation 5x^2-3x+1=10 ?
The solutions to the equation 5x^2-3x+1=10 are x=(3+3sqrt(21))/(10),x=(3-3sqrt(21))/(10)
Find the zeros of 5x^2-3x+1=10
The zeros of 5x^2-3x+1=10 are x=(3+3sqrt(21))/(10),x=(3-3sqrt(21))/(10)