What are the solutions to the equation 81-45x+4x^2=0 ?

The solutions to the equation 81-45x+4x^2=0 are x=9,x= 9/4

Find the zeros of 81-45x+4x^2=0

The zeros of 81-45x+4x^2=0 are x=9,x= 9/4

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81 - 45 x + 4 x^{2} = 0

x = 9, x = \frac{9}{4}

Decimal Notation

x = 9, x = 2.25

Solution steps

81 - 45 x + 4 x^{2} = 0

Write in the standard form

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

4 x^{2} - 45 x + 81 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 45 ) \pm ( - 45 ) ^{2} - 4 \cdot 4 \cdot 81}{2 \cdot 4}

(- 45)^{2} - 4 \cdot 4 \cdot 81 = 27

x_{1, 2} = \frac{- ( - 45 ) \pm 27}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 45 ) + 27}{2 \cdot 4}, x_{2} = \frac{- ( - 45 ) - 27}{2 \cdot 4}

x = \frac{- ( - 45 ) + 27}{2 \cdot 4} : 9

x = \frac{- ( - 45 ) - 27}{2 \cdot 4} : \frac{9}{4}

The solutions to the quadratic equation are:

x = 9, x = \frac{9}{4}

[5 h - 5] = 0

∣ 5 x + 2 ∣ > 12

\frac{1}{2} (4) (4)

\frac{- 1 + 9 i}{5 + 2 i}

\frac{x - 4}{x + 2} \geq 0

What are the solutions to the equation 81-45x+4x^2=0 ?
The solutions to the equation 81-45x+4x^2=0 are x=9,x= 9/4
Find the zeros of 81-45x+4x^2=0
The zeros of 81-45x+4x^2=0 are x=9,x= 9/4