What are the solutions to the equation 2x^2-3x-20=x^2+34 ?

The solutions to the equation 2x^2-3x-20=x^2+34 are x=9,x=-6

Find the zeros of 2x^2-3x-20=x^2+34

The zeros of 2x^2-3x-20=x^2+34 are x=9,x=-6

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2 x^{2} - 3 x - 20 = x^{2} + 34

x = 9, x = - 6

Solution steps

2 x^{2} - 3 x - 20 = x^{2} + 34

Move

34

to the left side

2 x^{2} - 3 x - 54 = x^{2}

Move

x^{2}

to the left side

x^{2} - 3 x - 54 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 54 )}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot (- 54) = 15

x_{1, 2} = \frac{- ( - 3 ) \pm 15}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 15}{2 \cdot 1}, x_{2} = \frac{- ( - 3 ) - 15}{2 \cdot 1}

x = \frac{- ( - 3 ) + 15}{2 \cdot 1} : 9

x = \frac{- ( - 3 ) - 15}{2 \cdot 1} : - 6

The solutions to the quadratic equation are:

x = 9, x = - 6

2 x^{2} + 2 = 10

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

x^{2} - 20 + 12 = 0

0 = 3 x (x - 1) - (3 x + 7)

10 x^{2} + 5 = 1

What are the solutions to the equation 2x^2-3x-20=x^2+34 ?
The solutions to the equation 2x^2-3x-20=x^2+34 are x=9,x=-6
Find the zeros of 2x^2-3x-20=x^2+34
The zeros of 2x^2-3x-20=x^2+34 are x=9,x=-6