What are the solutions to the equation 3x^2-8x=x^2+42 ?

The solutions to the equation 3x^2-8x=x^2+42 are x=7,x=-3

Find the zeros of 3x^2-8x=x^2+42

The zeros of 3x^2-8x=x^2+42 are x=7,x=-3

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3 x^{2} - 8 x = x^{2} + 42

x = 7, x = - 3

Solution steps

3 x^{2} - 8 x = x^{2} + 42

Move

42

to the left side

3 x^{2} - 8 x - 42 = x^{2}

Move

x^{2}

to the left side

2 x^{2} - 8 x - 42 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 2 ( - 42 )}{2 \cdot 2}

(- 8)^{2} - 4 \cdot 2 (- 42) = 20

x_{1, 2} = \frac{- ( - 8 ) \pm 20}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 8 ) + 20}{2 \cdot 2}, x_{2} = \frac{- ( - 8 ) - 20}{2 \cdot 2}

x = \frac{- ( - 8 ) + 20}{2 \cdot 2} : 7

x = \frac{- ( - 8 ) - 20}{2 \cdot 2} : - 3

The solutions to the quadratic equation are:

x = 7, x = - 3

a^{2} + 3^{2} = 5^{2}

3 (x + 1)^{2} = 6 x + 12

- 3 x^{2} + 4 x - 1 = 0

y (y + 6) = 2

x^{2} + 1 = - 8

What are the solutions to the equation 3x^2-8x=x^2+42 ?
The solutions to the equation 3x^2-8x=x^2+42 are x=7,x=-3
Find the zeros of 3x^2-8x=x^2+42
The zeros of 3x^2-8x=x^2+42 are x=7,x=-3