What are the solutions to the equation 4(10-x)^2=5x^2 ?

The solutions to the equation 4(10-x)^2=5x^2 are x=-20(2+sqrt(5)),x=20(sqrt(5)-2)

Find the zeros of 4(10-x)^2=5x^2

The zeros of 4(10-x)^2=5x^2 are x=-20(2+sqrt(5)),x=20(sqrt(5)-2)

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4 (10 - x)^{2} = 5 x^{2}

x = - 20 (2 + 5), x = 20 (5 - 2)

Decimal Notation

x = - 84.72135 \dots, x = 4.72135 \dots

Solution steps

4 (10 - x)^{2} = 5 x^{2}

Expand

4 (10 - x)^{2} : 400 - 80 x + 4 x^{2}

400 - 80 x + 4 x^{2} = 5 x^{2}

Move

5 x^{2}

to the left side

400 - 80 x - x^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

- x^{2} - 80 x + 400 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 80 ) \pm ( - 80 ) ^{2} - 4 ( - 1 ) \cdot 400}{2 ( - 1 )}

(- 80)^{2} - 4 (- 1) \cdot 400 = 405

x_{1, 2} = \frac{- ( - 80 ) \pm 40 5}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- ( - 80 ) + 40 5}{2 ( - 1 )}, x_{2} = \frac{- ( - 80 ) - 40 5}{2 ( - 1 )}

x = \frac{- ( - 80 ) + 40 5}{2 ( - 1 )} : - 20 (2 + 5)

x = \frac{- ( - 80 ) - 40 5}{2 ( - 1 )} : 20 (5 - 2)

The solutions to the quadratic equation are:

x = - 20 (2 + 5), x = 20 (5 - 2)

x, 4 x^{2} - y^{2} + 4 z^{2} - 4 = 0

(5 x + 3) (2 x - 8) = 0

2 x^{2} + 3 = x^{2} - 2 x + 4

t, f = (5 t + 2) (t - 2)

\frac{n ( n + 1 )}{2} = 100

What are the solutions to the equation 4(10-x)^2=5x^2 ?
The solutions to the equation 4(10-x)^2=5x^2 are x=-20(2+sqrt(5)),x=20(sqrt(5)-2)
Find the zeros of 4(10-x)^2=5x^2
The zeros of 4(10-x)^2=5x^2 are x=-20(2+sqrt(5)),x=20(sqrt(5)-2)