What are the solutions to the equation 2x+(2x^2+x)=20 ?

The solutions to the equation 2x+(2x^2+x)=20 are x= 5/2 ,x=-4

Find the zeros of 2x+(2x^2+x)=20

The zeros of 2x+(2x^2+x)=20 are x= 5/2 ,x=-4

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2 x + (2 x^{2} + x) = 20

x = \frac{5}{2}, x = - 4

Decimal Notation

x = 2.5, x = - 4

Solution steps

2 x + (2 x^{2} + x) = 20

Refine

2 x^{2} + 3 x = 20

Move

20

to the left side

2 x^{2} + 3 x - 20 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 2 ( - 20 )}{2 \cdot 2}

3^{2} - 4 \cdot 2 (- 20) = 13

x_{1, 2} = \frac{- 3 \pm 13}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 3 + 13}{2 \cdot 2}, x_{2} = \frac{- 3 - 13}{2 \cdot 2}

x = \frac{- 3 + 13}{2 \cdot 2} : \frac{5}{2}

x = \frac{- 3 - 13}{2 \cdot 2} : - 4

The solutions to the quadratic equation are:

x = \frac{5}{2}, x = - 4

(3 x + 2)^{2} = 81

n^{1} + n^{2} = 4, 0 = 24

3 x^{2} - 5 x - 2 = 0

2 x^{2} - 13 x - 15 = 0

(x - 1296)^{2} = 0

What are the solutions to the equation 2x+(2x^2+x)=20 ?
The solutions to the equation 2x+(2x^2+x)=20 are x= 5/2 ,x=-4
Find the zeros of 2x+(2x^2+x)=20
The zeros of 2x+(2x^2+x)=20 are x= 5/2 ,x=-4