What are the solutions to the equation x^2+2x+5x+10-5x+10=28 ?

The solutions to the equation x^2+2x+5x+10-5x+10=28 are x=2,x=-4

Find the zeros of x^2+2x+5x+10-5x+10=28

The zeros of x^2+2x+5x+10-5x+10=28 are x=2,x=-4

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x^{2} + 2 x + 5 x + 10 - 5 x + 10 = 28

x = 2, x = - 4

Solution steps

x^{2} + 2 x + 5 x + 10 - 5 x + 10 = 28

Refine

x^{2} + 2 x + 20 = 28

Move

28

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

180 = - 16 t^{2} + 120 t

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

7 x^{2} - 6 x + 5 = - 4

36 x^{2} = 9

18 x^{2} + 50 = 0

What are the solutions to the equation x^2+2x+5x+10-5x+10=28 ?
The solutions to the equation x^2+2x+5x+10-5x+10=28 are x=2,x=-4
Find the zeros of x^2+2x+5x+10-5x+10=28
The zeros of x^2+2x+5x+10-5x+10=28 are x=2,x=-4