What are the solutions to the equation 4x^2-x-48=3x^2-12+3x ?

The solutions to the equation 4x^2-x-48=3x^2-12+3x are x=2(1+sqrt(10)),x=2(1-sqrt(10))

Find the zeros of 4x^2-x-48=3x^2-12+3x

The zeros of 4x^2-x-48=3x^2-12+3x are x=2(1+sqrt(10)),x=2(1-sqrt(10))

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4 x^{2} - x - 48 = 3 x^{2} - 12 + 3 x

x = 2 (1 + 10), x = 2 (1 - 10)

Decimal Notation

x = 8.32455 \dots, x = - 4.32455 \dots

Solution steps

4 x^{2} - x - 48 = 3 x^{2} - 12 + 3 x

Move

3 x

to the left side

4 x^{2} - 4 x - 48 = 3 x^{2} - 12

Move

12

to the left side

4 x^{2} - 4 x - 36 = 3 x^{2}

Move

3 x^{2}

to the left side

x^{2} - 4 x - 36 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot ( - 36 )}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot (- 36) = 410

x_{1, 2} = \frac{- ( - 4 ) \pm 4 10}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 4 10}{2 \cdot 1}, x_{2} = \frac{- ( - 4 ) - 4 10}{2 \cdot 1}

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

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

The solutions to the quadratic equation are:

x = 2 (1 + 10), x = 2 (1 - 10)

1.5 \cdot 5

lo g_{10} (- 1)

3 x + 15 = 0

\frac{10 !}{6 ! \cdot 4 !}

0.8 \cdot 60

What are the solutions to the equation 4x^2-x-48=3x^2-12+3x ?
The solutions to the equation 4x^2-x-48=3x^2-12+3x are x=2(1+sqrt(10)),x=2(1-sqrt(10))
Find the zeros of 4x^2-x-48=3x^2-12+3x
The zeros of 4x^2-x-48=3x^2-12+3x are x=2(1+sqrt(10)),x=2(1-sqrt(10))