What are the solutions to the equation 2x^2-180x+400=0 ?

The solutions to the equation 2x^2-180x+400=0 are x=5(9+sqrt(73)),x=5(9-sqrt(73))

Find the zeros of 2x^2-180x+400=0

The zeros of 2x^2-180x+400=0 are x=5(9+sqrt(73)),x=5(9-sqrt(73))

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2 x^{2} - 180 x + 400 = 0

x = 5 (9 + 73), x = 5 (9 - 73)

Decimal Notation

x = 87.72001 \dots, x = 2.27998 \dots

Solution steps

2 x^{2} - 180 x + 400 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 180 ) \pm ( - 180 ) ^{2} - 4 \cdot 2 \cdot 400}{2 \cdot 2}

(- 180)^{2} - 4 \cdot 2 \cdot 400 = 2073

x_{1, 2} = \frac{- ( - 180 ) \pm 20 73}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 180 ) + 20 73}{2 \cdot 2}, x_{2} = \frac{- ( - 180 ) - 20 73}{2 \cdot 2}

x = \frac{- ( - 180 ) + 20 73}{2 \cdot 2} : 5 (9 + 73)

x = \frac{- ( - 180 ) - 20 73}{2 \cdot 2} : 5 (9 - 73)

The solutions to the quadratic equation are:

x = 5 (9 + 73), x = 5 (9 - 73)

(2 x + 5) (3 x - 2) = (x + 1)

x^{2} + 10 x = - 15

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

z^{2} - 2 z - 24 = 0

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

What are the solutions to the equation 2x^2-180x+400=0 ?
The solutions to the equation 2x^2-180x+400=0 are x=5(9+sqrt(73)),x=5(9-sqrt(73))
Find the zeros of 2x^2-180x+400=0
The zeros of 2x^2-180x+400=0 are x=5(9+sqrt(73)),x=5(9-sqrt(73))