What are the solutions to the equation 2x^2-16x=12x+48 ?

The solutions to the equation 2x^2-16x=12x+48 are x=7+sqrt(73),x=7-sqrt(73)

Find the zeros of 2x^2-16x=12x+48

The zeros of 2x^2-16x=12x+48 are x=7+sqrt(73),x=7-sqrt(73)

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2 x^{2} - 16 x = 12 x + 48

x = 7 + 73, x = 7 - 73

Decimal Notation

x = 15.54400 \dots, x = - 1.54400 \dots

Solution steps

2 x^{2} - 16 x = 12 x + 48

Move

48

to the left side

2 x^{2} - 16 x - 48 = 12 x

Move

12 x

to the left side

2 x^{2} - 28 x - 48 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 28 ) \pm ( - 28 ) ^{2} - 4 \cdot 2 ( - 48 )}{2 \cdot 2}

(- 28)^{2} - 4 \cdot 2 (- 48) = 473

x_{1, 2} = \frac{- ( - 28 ) \pm 4 73}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 28 ) + 4 73}{2 \cdot 2}, x_{2} = \frac{- ( - 28 ) - 4 73}{2 \cdot 2}

x = \frac{- ( - 28 ) + 4 73}{2 \cdot 2} : 7 + 73

x = \frac{- ( - 28 ) - 4 73}{2 \cdot 2} : 7 - 73

The solutions to the quadratic equation are:

x = 7 + 73, x = 7 - 73

(\frac{5 x - 1}{3})^{2} = \frac{4}{9}

x^{2} + 2 (m + 1) x - 3 m (m - 2) = 0

x, f = f (x^{2}) - 2 f (1)

x, x^{2} + y^{2} + z^{2} - 25 = 0

- x^{2} + 10 x + 24 = 0

What are the solutions to the equation 2x^2-16x=12x+48 ?
The solutions to the equation 2x^2-16x=12x+48 are x=7+sqrt(73),x=7-sqrt(73)
Find the zeros of 2x^2-16x=12x+48
The zeros of 2x^2-16x=12x+48 are x=7+sqrt(73),x=7-sqrt(73)