What are the solutions to the equation 81x=4x^2-20x+25 ?

The solutions to the equation 81x=4x^2-20x+25 are x=25,x= 1/4

Find the zeros of 81x=4x^2-20x+25

The zeros of 81x=4x^2-20x+25 are x=25,x= 1/4

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81 x = 4 x^{2} - 20 x + 25

x = 25, x = \frac{1}{4}

Decimal Notation

x = 25, x = 0.25

Solution steps

81 x = 4 x^{2} - 20 x + 25

Switch sides

4 x^{2} - 20 x + 25 = 81 x

Move

81 x

to the left side

4 x^{2} - 101 x + 25 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 101 ) \pm ( - 101 ) ^{2} - 4 \cdot 4 \cdot 25}{2 \cdot 4}

(- 101)^{2} - 4 \cdot 4 \cdot 25 = 99

x_{1, 2} = \frac{- ( - 101 ) \pm 99}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 101 ) + 99}{2 \cdot 4}, x_{2} = \frac{- ( - 101 ) - 99}{2 \cdot 4}

x = \frac{- ( - 101 ) + 99}{2 \cdot 4} : 25

x = \frac{- ( - 101 ) - 99}{2 \cdot 4} : \frac{1}{4}

The solutions to the quadratic equation are:

x = 25, x = \frac{1}{4}

(2 x + 3) (4 x + 1) = 0

(x - 9) (x + 1) = 3 x

3 c^{2} + 9 c - 7 = - 6

10 a^{2} - 3 a - 1 = 0

x^{2} - 8 x - 29 = 0

What are the solutions to the equation 81x=4x^2-20x+25 ?
The solutions to the equation 81x=4x^2-20x+25 are x=25,x= 1/4
Find the zeros of 81x=4x^2-20x+25
The zeros of 81x=4x^2-20x+25 are x=25,x= 1/4