What are the solutions to the equation 2x^2-(x-3)(x+5)=7(x+3) ?

The solutions to the equation 2x^2-(x-3)(x+5)=7(x+3) are x=(9+sqrt(105))/2 ,x=(9-sqrt(105))/2

Find the zeros of 2x^2-(x-3)(x+5)=7(x+3)

The zeros of 2x^2-(x-3)(x+5)=7(x+3) are x=(9+sqrt(105))/2 ,x=(9-sqrt(105))/2

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2 x^{2} - (x - 3) (x + 5) = 7 (x + 3)

x = \frac{9 + 105}{2}, x = \frac{9 - 105}{2}

Solution steps

2 x^{2} - (x - 3) (x + 5) = 7 (x + 3)

Expand

2 x^{2} - (x - 3) (x + 5) : x^{2} - 2 x + 15

Expand

7 (x + 3) : 7 x + 21

x^{2} - 2 x + 15 = 7 x + 21

Move

21

to the left side

x^{2} - 2 x - 6 = 7 x

Move

7 x

to the left side

x^{2} - 9 x - 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 1 \cdot ( - 6 )}{2 \cdot 1}

(- 9)^{2} - 4 \cdot 1 \cdot (- 6) = 105

x_{1, 2} = \frac{- ( - 9 ) \pm 105}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 9 ) + 105}{2 \cdot 1}, x_{2} = \frac{- ( - 9 ) - 105}{2 \cdot 1}

x = \frac{- ( - 9 ) + 105}{2 \cdot 1} : \frac{9 + 105}{2}

x = \frac{- ( - 9 ) - 105}{2 \cdot 1} : \frac{9 - 105}{2}

The solutions to the quadratic equation are:

x = \frac{9 + 105}{2}, x = \frac{9 - 105}{2}

- 6 x y^{5} (16 x^{4} y)

3 x^{2} + 5 x - 1 \leq 5 x^{2} - 6

- 2.5 x^{2} + 400 x + 3600 = 0

20 x^{2} - 17 x - 63

3 i + 2 i

What are the solutions to the equation 2x^2-(x-3)(x+5)=7(x+3) ?
The solutions to the equation 2x^2-(x-3)(x+5)=7(x+3) are x=(9+sqrt(105))/2 ,x=(9-sqrt(105))/2
Find the zeros of 2x^2-(x-3)(x+5)=7(x+3)
The zeros of 2x^2-(x-3)(x+5)=7(x+3) are x=(9+sqrt(105))/2 ,x=(9-sqrt(105))/2