What are the solutions to the equation 4x^2-36x+81=0 ?

The solutions to the equation 4x^2-36x+81=0 are x= 9/2

Find the zeros of 4x^2-36x+81=0

The zeros of 4x^2-36x+81=0 are x= 9/2

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4 x^{2} - 36 x + 81 = 0

x = \frac{9}{2}

Decimal Notation

x = 4.5

Solution steps

4 x^{2} - 36 x + 81 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 36 ) \pm ( - 36 ) ^{2} - 4 \cdot 4 \cdot 81}{2 \cdot 4}

(- 36)^{2} - 4 \cdot 4 \cdot 81 = 0

x_{1, 2} = \frac{- ( - 36 ) \pm 0}{2 \cdot 4}

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

\frac{- ( - 36 )}{2 \cdot 4} = \frac{9}{2}

x = \frac{9}{2}

The solution to the quadratic equation is:

x = \frac{9}{2}

x^{2} (7 - 6 x) \leq 1

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

\frac{7800}{0.15 - 0.05}

2 n - 5 n + 16 = - 11

3 x^{2} + 3 x = 5

What are the solutions to the equation 4x^2-36x+81=0 ?
The solutions to the equation 4x^2-36x+81=0 are x= 9/2
Find the zeros of 4x^2-36x+81=0
The zeros of 4x^2-36x+81=0 are x= 9/2