What are the solutions to the equation 5x^2-20x+3=0 ?

The solutions to the equation 5x^2-20x+3=0 are x=(10+sqrt(85))/5 ,x=(10-sqrt(85))/5

Find the zeros of 5x^2-20x+3=0

The zeros of 5x^2-20x+3=0 are x=(10+sqrt(85))/5 ,x=(10-sqrt(85))/5

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5 x^{2} - 20 x + 3 = 0

x = \frac{10 + 85}{5}, x = \frac{10 - 85}{5}

Decimal Notation

x = 3.84390 \dots, x = 0.15609 \dots

Solution steps

5 x^{2} - 20 x + 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 \cdot 5 \cdot 3}{2 \cdot 5}

(- 20)^{2} - 4 \cdot 5 \cdot 3 = 285

x_{1, 2} = \frac{- ( - 20 ) \pm 2 85}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 20 ) + 2 85}{2 \cdot 5}, x_{2} = \frac{- ( - 20 ) - 2 85}{2 \cdot 5}

x = \frac{- ( - 20 ) + 2 85}{2 \cdot 5} : \frac{10 + 85}{5}

x = \frac{- ( - 20 ) - 2 85}{2 \cdot 5} : \frac{10 - 85}{5}

The solutions to the quadratic equation are:

x = \frac{10 + 85}{5}, x = \frac{10 - 85}{5}

10 x^{2} + 2 x y - 15 x - 3 y

\frac{m ^{\frac{2}{3}} n ^{2}}{m ^{- 1} n ^{3}}

7 (\frac{1}{2} x + 2) - 4 < - 5

36 a^{4} - 25 b^{4}

(x - 2) (x + 4) = 0

What are the solutions to the equation 5x^2-20x+3=0 ?
The solutions to the equation 5x^2-20x+3=0 are x=(10+sqrt(85))/5 ,x=(10-sqrt(85))/5
Find the zeros of 5x^2-20x+3=0
The zeros of 5x^2-20x+3=0 are x=(10+sqrt(85))/5 ,x=(10-sqrt(85))/5