What are the solutions to the equation 4x^2+48x=288 ?

The solutions to the equation 4x^2+48x=288 are x=6(sqrt(3)-1),x=-6(1+sqrt(3))

Find the zeros of 4x^2+48x=288

The zeros of 4x^2+48x=288 are x=6(sqrt(3)-1),x=-6(1+sqrt(3))

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4 x^{2} + 48 x = 288

x = 6 (3 - 1), x = - 6 (1 + 3)

Decimal Notation

x = 4.39230 \dots, x = - 16.39230 \dots

Solution steps

4 x^{2} + 48 x = 288

Move

288

to the left side

4 x^{2} + 48 x - 288 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 48 \pm 4 8 ^{2} - 4 \cdot 4 ( - 288 )}{2 \cdot 4}

4 8^{2} - 4 \cdot 4 (- 288) = 483

x_{1, 2} = \frac{- 48 \pm 48 3}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 48 + 48 3}{2 \cdot 4}, x_{2} = \frac{- 48 - 48 3}{2 \cdot 4}

x = \frac{- 48 + 48 3}{2 \cdot 4} : 6 (3 - 1)

x = \frac{- 48 - 48 3}{2 \cdot 4} : - 6 (1 + 3)

The solutions to the quadratic equation are:

x = 6 (3 - 1), x = - 6 (1 + 3)

4 (x - 1)^{2} = 25

8 x^{2} + 9 x = 0

x^{2} - 12 x - 27 = 0

- x^{2} + 35 = 19

- \frac{3}{4} x^{2} + \frac{9}{4} x + 3 = 3

What are the solutions to the equation 4x^2+48x=288 ?
The solutions to the equation 4x^2+48x=288 are x=6(sqrt(3)-1),x=-6(1+sqrt(3))
Find the zeros of 4x^2+48x=288
The zeros of 4x^2+48x=288 are x=6(sqrt(3)-1),x=-6(1+sqrt(3))