What are the solutions to the equation 20x^2-41x+20=0 ?

The solutions to the equation 20x^2-41x+20=0 are x= 5/4 ,x= 4/5

Find the zeros of 20x^2-41x+20=0

The zeros of 20x^2-41x+20=0 are x= 5/4 ,x= 4/5

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20 x^{2} - 41 x + 20 = 0

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

Decimal Notation

x = 1.25, x = 0.8

Solution steps

20 x^{2} - 41 x + 20 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 41 ) \pm ( - 41 ) ^{2} - 4 \cdot 20 \cdot 20}{2 \cdot 20}

(- 41)^{2} - 4 \cdot 20 \cdot 20 = 9

x_{1, 2} = \frac{- ( - 41 ) \pm 9}{2 \cdot 20}

Separate the solutions

x_{1} = \frac{- ( - 41 ) + 9}{2 \cdot 20}, x_{2} = \frac{- ( - 41 ) - 9}{2 \cdot 20}

x = \frac{- ( - 41 ) + 9}{2 \cdot 20} : \frac{5}{4}

x = \frac{- ( - 41 ) - 9}{2 \cdot 20} : \frac{4}{5}

The solutions to the quadratic equation are:

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

x, x^{2} + 8 x y - 4 y^{2} + 3 x + 8 y + 7 = 0

4 X^{2} = 81

2 y^{2} - 3 = 0

x^{2} + 18 x + 81 = 25 + 81

x^{2} - 12 x = 0

What are the solutions to the equation 20x^2-41x+20=0 ?
The solutions to the equation 20x^2-41x+20=0 are x= 5/4 ,x= 4/5
Find the zeros of 20x^2-41x+20=0
The zeros of 20x^2-41x+20=0 are x= 5/4 ,x= 4/5