What are the solutions to the equation x^2+3+4+2x=15 ?

The solutions to the equation x^2+3+4+2x=15 are x=2,x=-4

Find the zeros of x^2+3+4+2x=15

The zeros of x^2+3+4+2x=15 are x=2,x=-4

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x^{2} + 3 + 4 + 2 x = 15

x = 2, x = - 4

Solution steps

x^{2} + 3 + 4 + 2 x = 15

Refine

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

Move

15

to the left side

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

Solve with the quadratic formula

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

2^{2} - 4 \cdot 1 \cdot (- 8) = 6

x_{1, 2} = \frac{- 2 \pm 6}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 2 + 6}{2 \cdot 1}, x_{2} = \frac{- 2 - 6}{2 \cdot 1}

x = \frac{- 2 + 6}{2 \cdot 1} : 2

x = \frac{- 2 - 6}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

x = 2, x = - 4

- 6 x^{2} - 96 x + 355 = - 12 x + 1

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

x^{2} - 500 = 125

x^{2} + 2 x - 48 = 0

(x - 5)^{2} - (x - 5) - 30 = 0

What are the solutions to the equation x^2+3+4+2x=15 ?
The solutions to the equation x^2+3+4+2x=15 are x=2,x=-4
Find the zeros of x^2+3+4+2x=15
The zeros of x^2+3+4+2x=15 are x=2,x=-4