What are the solutions to the equation 27x^2+5=48x ?

The solutions to the equation 27x^2+5=48x are x= 5/3 ,x= 1/9

Find the zeros of 27x^2+5=48x

The zeros of 27x^2+5=48x are x= 5/3 ,x= 1/9

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27 x^{2} + 5 = 48 x

x = \frac{5}{3}, x = \frac{1}{9}

Decimal

x = 1.66666 \dots, x = 0.11111 \dots

Solution steps

27 x^{2} + 5 = 48 x

Move

48 x

to the left side

27 x^{2} + 5 - 48 x = 0

Write in the standard form

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

27 x^{2} - 48 x + 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 48 ) \pm ( - 48 ) ^{2} - 4 \cdot 27 \cdot 5}{2 \cdot 27}

(- 48)^{2} - 4 \cdot 27 \cdot 5 = 42

x_{1, 2} = \frac{- ( - 48 ) \pm 42}{2 \cdot 27}

Separate the solutions

x_{1} = \frac{- ( - 48 ) + 42}{2 \cdot 27}, x_{2} = \frac{- ( - 48 ) - 42}{2 \cdot 27}

x = \frac{- ( - 48 ) + 42}{2 \cdot 27} : \frac{5}{3}

x = \frac{- ( - 48 ) - 42}{2 \cdot 27} : \frac{1}{9}

The solutions to the quadratic equation are:

x = \frac{5}{3}, x = \frac{1}{9}

3 x^{2} - 2 = 1

x^{2} - x + 1 = 1

x^{2} - 14 x + 9 = 0

y^{2} = 8 y + 20

(x - 21)^{2} = 28

What are the solutions to the equation 27x^2+5=48x ?
The solutions to the equation 27x^2+5=48x are x= 5/3 ,x= 1/9
Find the zeros of 27x^2+5=48x
The zeros of 27x^2+5=48x are x= 5/3 ,x= 1/9