What are the solutions to the equation (60-3x)\times (32-x)=480 ?

The solutions to the equation (60-3x)\times (32-x)=480 are x=40,x=12

Find the zeros of (60-3x)\times (32-x)=480

The zeros of (60-3x)\times (32-x)=480 are x=40,x=12

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(60 - 3 x) \cdot (32 - x) = 480

x = 40, x = 12

Solution steps

(60 - 3 x) (32 - x) = 480

Expand

(60 - 3 x) (32 - x) : 1920 - 156 x + 3 x^{2}

1920 - 156 x + 3 x^{2} = 480

Move

480

to the left side

3 x^{2} - 156 x + 1440 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 156 ) \pm ( - 156 ) ^{2} - 4 \cdot 3 \cdot 1440}{2 \cdot 3}

(- 156)^{2} - 4 \cdot 3 \cdot 1440 = 84

x_{1, 2} = \frac{- ( - 156 ) \pm 84}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 156 ) + 84}{2 \cdot 3}, x_{2} = \frac{- ( - 156 ) - 84}{2 \cdot 3}

x = \frac{- ( - 156 ) + 84}{2 \cdot 3} : 40

x = \frac{- ( - 156 ) - 84}{2 \cdot 3} : 12

The solutions to the quadratic equation are:

x = 40, x = 12

- x^{2} = 12

x, x^{2} + y^{2} = c^{2}

4 a^{2} x^{2} + 8 a x + 3 = 0

2 x^{2} - 11 x + 16 = 0

x, 3 k x^{2} - 5 x = 4 (x - 1)

What are the solutions to the equation (60-3x)\times (32-x)=480 ?
The solutions to the equation (60-3x)\times (32-x)=480 are x=40,x=12
Find the zeros of (60-3x)\times (32-x)=480
The zeros of (60-3x)\times (32-x)=480 are x=40,x=12