What are the solutions to the equation 2x+3(x-4)^2=37+(x-3)(x+3) ?

The solutions to the equation 2x+3(x-4)^2=37+(x-3)(x+3) are x=10,x=1

Find the zeros of 2x+3(x-4)^2=37+(x-3)(x+3)

The zeros of 2x+3(x-4)^2=37+(x-3)(x+3) are x=10,x=1

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2 x + 3 (x - 4)^{2} = 37 + (x - 3) (x + 3)

x = 10, x = 1

Solution steps

2 x + 3 (x - 4)^{2} = 37 + (x - 3) (x + 3)

Expand

2 x + 3 (x - 4)^{2} : 3 x^{2} - 22 x + 48

Expand

37 + (x - 3) (x + 3) : x^{2} + 28

3 x^{2} - 22 x + 48 = x^{2} + 28

Move

28

to the left side

3 x^{2} - 22 x + 20 = x^{2}

Move

x^{2}

to the left side

2 x^{2} - 22 x + 20 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 22 ) \pm ( - 22 ) ^{2} - 4 \cdot 2 \cdot 20}{2 \cdot 2}

(- 22)^{2} - 4 \cdot 2 \cdot 20 = 18

x_{1, 2} = \frac{- ( - 22 ) \pm 18}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 22 ) + 18}{2 \cdot 2}, x_{2} = \frac{- ( - 22 ) - 18}{2 \cdot 2}

x = \frac{- ( - 22 ) + 18}{2 \cdot 2} : 10

x = \frac{- ( - 22 ) - 18}{2 \cdot 2} : 1

The solutions to the quadratic equation are:

x = 10, x = 1

5175

2 x^{2} - 4 x + 5 = 0

6 g^{3} + 18 g^{2} + 5 g + 15

1.3 \cdot 2

2 ∣ 25 x ∣ + 50 = 100

What are the solutions to the equation 2x+3(x-4)^2=37+(x-3)(x+3) ?
The solutions to the equation 2x+3(x-4)^2=37+(x-3)(x+3) are x=10,x=1
Find the zeros of 2x+3(x-4)^2=37+(x-3)(x+3)
The zeros of 2x+3(x-4)^2=37+(x-3)(x+3) are x=10,x=1