What are the solutions to the equation x^2-9x+81/4 = 1/4 ?

The solutions to the equation x^2-9x+81/4 = 1/4 are x=5,x=4

Find the zeros of x^2-9x+81/4 = 1/4

The zeros of x^2-9x+81/4 = 1/4 are x=5,x=4

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x^{2} - 9 x + \frac{81}{4} = \frac{1}{4}

x = 5, x = 4

Solution steps

x^{2} - 9 x + \frac{81}{4} = \frac{1}{4}

Multiply both sides by

4

4 x^{2} - 36 x + 81 = 1

Move

1

to the left side

4 x^{2} - 36 x + 80 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 36 ) \pm ( - 36 ) ^{2} - 4 \cdot 4 \cdot 80}{2 \cdot 4}

(- 36)^{2} - 4 \cdot 4 \cdot 80 = 4

x_{1, 2} = \frac{- ( - 36 ) \pm 4}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 36 ) + 4}{2 \cdot 4}, x_{2} = \frac{- ( - 36 ) - 4}{2 \cdot 4}

x = \frac{- ( - 36 ) + 4}{2 \cdot 4} : 5

x = \frac{- ( - 36 ) - 4}{2 \cdot 4} : 4

The solutions to the quadratic equation are:

x = 5, x = 4

0.5 x + 6 = 2 (2 x - 3) x

x, f = 3 x^{2} yz - x y^{2} z^{2}

8 x^{2} = 12 - 4 x

s^{2} - 2 s + 10

x, 5 x^{2} - 3 x y + 5 y^{2} - 27 = 0

What are the solutions to the equation x^2-9x+81/4 = 1/4 ?
The solutions to the equation x^2-9x+81/4 = 1/4 are x=5,x=4
Find the zeros of x^2-9x+81/4 = 1/4
The zeros of x^2-9x+81/4 = 1/4 are x=5,x=4