What are the solutions to the equation (x^2}{12}-\frac{4x)/3+5=0 ?

The solutions to the equation (x^2}{12}-\frac{4x)/3+5=0 are x=10,x=6

Find the zeros of (x^2}{12}-\frac{4x)/3+5=0

The zeros of (x^2}{12}-\frac{4x)/3+5=0 are x=10,x=6

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\frac{x ^{2}}{12} - \frac{4 x}{3} + 5 = 0

x = 10, x = 6

Solution steps

\frac{x ^{2}}{12} - \frac{4 x}{3} + 5 = 0

Find Least Common Multiplier of

12, 3 : 12

Multiply by LCM=

12

\frac{x ^{2}}{12} \cdot 12 - \frac{4 x}{3} \cdot 12 + 5 \cdot 12 = 0 \cdot 12

Simplify

x^{2} - 16 x + 60 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 1 \cdot 60}{2 \cdot 1}

(- 16)^{2} - 4 \cdot 1 \cdot 60 = 4

x_{1, 2} = \frac{- ( - 16 ) \pm 4}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 16 ) + 4}{2 \cdot 1}, x_{2} = \frac{- ( - 16 ) - 4}{2 \cdot 1}

x = \frac{- ( - 16 ) + 4}{2 \cdot 1} : 10

x = \frac{- ( - 16 ) - 4}{2 \cdot 1} : 6

The solutions to the quadratic equation are:

x = 10, x = 6

7 x + 0 = 2 x + 18 + 2 x + 18

6 x - 4 + 2 x = 4 (5 + 2 x)

(5 r^{3})^{\frac{1}{4}}

∣ x + 4 ∣ = 0

a = \frac{1}{2} π (12.9)^{2}

What are the solutions to the equation (x^2}{12}-\frac{4x)/3+5=0 ?
The solutions to the equation (x^2}{12}-\frac{4x)/3+5=0 are x=10,x=6
Find the zeros of (x^2}{12}-\frac{4x)/3+5=0
The zeros of (x^2}{12}-\frac{4x)/3+5=0 are x=10,x=6