What are the solutions to the equation 5x^2-28x+19=0 ?

The solutions to the equation 5x^2-28x+19=0 are x=(14+sqrt(101))/5 ,x=(14-sqrt(101))/5

Find the zeros of 5x^2-28x+19=0

The zeros of 5x^2-28x+19=0 are x=(14+sqrt(101))/5 ,x=(14-sqrt(101))/5

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5 x^{2} - 28 x + 19 = 0

x = \frac{14 + 101}{5}, x = \frac{14 - 101}{5}

Decimal Notation

x = 4.80997 \dots, x = 0.79002 \dots

Solution steps

5 x^{2} - 28 x + 19 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 28 ) \pm ( - 28 ) ^{2} - 4 \cdot 5 \cdot 19}{2 \cdot 5}

(- 28)^{2} - 4 \cdot 5 \cdot 19 = 2101

x_{1, 2} = \frac{- ( - 28 ) \pm 2 101}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 28 ) + 2 101}{2 \cdot 5}, x_{2} = \frac{- ( - 28 ) - 2 101}{2 \cdot 5}

x = \frac{- ( - 28 ) + 2 101}{2 \cdot 5} : \frac{14 + 101}{5}

x = \frac{- ( - 28 ) - 2 101}{2 \cdot 5} : \frac{14 - 101}{5}

The solutions to the quadratic equation are:

x = \frac{14 + 101}{5}, x = \frac{14 - 101}{5}

3 x^{2} - 5 = 4

4 x^{2} - 108 = 0

- x^{2} - 2 x + 9 = 0

x^{2} - 5 x + 8 = 2 x + 4

x^{2} - x + 1 = 7

What are the solutions to the equation 5x^2-28x+19=0 ?
The solutions to the equation 5x^2-28x+19=0 are x=(14+sqrt(101))/5 ,x=(14-sqrt(101))/5
Find the zeros of 5x^2-28x+19=0
The zeros of 5x^2-28x+19=0 are x=(14+sqrt(101))/5 ,x=(14-sqrt(101))/5