What are the solutions to the equation 54=0.04x^2+0.2x+12 ?

The solutions to the equation 54=0.04x^2+0.2x+12 are x=30,x=-35

Find the zeros of 54=0.04x^2+0.2x+12

The zeros of 54=0.04x^2+0.2x+12 are x=30,x=-35

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54 = 0.04 x^{2} + 0.2 x + 12

x = 30, x = - 35

Solution steps

54 = 0.04 x^{2} + 0.2 x + 12

Multiply both sides by

100

5400 = 4 x^{2} + 20 x + 1200

Switch sides

4 x^{2} + 20 x + 1200 = 5400

Move

5400

to the left side

4 x^{2} + 20 x - 4200 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 4 ( - 4200 )}{2 \cdot 4}

2 0^{2} - 4 \cdot 4 (- 4200) = 260

x_{1, 2} = \frac{- 20 \pm 260}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 20 + 260}{2 \cdot 4}, x_{2} = \frac{- 20 - 260}{2 \cdot 4}

x = \frac{- 20 + 260}{2 \cdot 4} : 30

x = \frac{- 20 - 260}{2 \cdot 4} : - 35

The solutions to the quadratic equation are:

x = 30, x = - 35

x^{2} - 6 x - 2

x^{2} - 6 x - 6

t^{2} - 4 t - 2 = 0

5 x^{2} = 18 x + 8

2 x^{2} = - 2 x - 5

What are the solutions to the equation 54=0.04x^2+0.2x+12 ?
The solutions to the equation 54=0.04x^2+0.2x+12 are x=30,x=-35
Find the zeros of 54=0.04x^2+0.2x+12
The zeros of 54=0.04x^2+0.2x+12 are x=30,x=-35