What are the solutions to the equation 15=7x+2x^2 ?

The solutions to the equation 15=7x+2x^2 are x= 3/2 ,x=-5

Find the zeros of 15=7x+2x^2

The zeros of 15=7x+2x^2 are x= 3/2 ,x=-5

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15 = 7 x + 2 x^{2}

x = \frac{3}{2}, x = - 5

Decimal Notation

x = 1.5, x = - 5

Solution steps

15 = 7 x + 2 x^{2}

Switch sides

7 x + 2 x^{2} = 15

Move

15

to the left side

7 x + 2 x^{2} - 15 = 0

Write in the standard form

a x^{2} + b x + c = 0

2 x^{2} + 7 x - 15 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 2 ( - 15 )}{2 \cdot 2}

7^{2} - 4 \cdot 2 (- 15) = 13

x_{1, 2} = \frac{- 7 \pm 13}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 7 + 13}{2 \cdot 2}, x_{2} = \frac{- 7 - 13}{2 \cdot 2}

x = \frac{- 7 + 13}{2 \cdot 2} : \frac{3}{2}

x = \frac{- 7 - 13}{2 \cdot 2} : - 5

The solutions to the quadratic equation are:

x = \frac{3}{2}, x = - 5

3 (5 - 2 x) = 10 - x

r^{2} + 8 r + 20 = 0

2 x - 2 > 3 x - 8

5 + 3 x = 0

2^{- 7}

What are the solutions to the equation 15=7x+2x^2 ?
The solutions to the equation 15=7x+2x^2 are x= 3/2 ,x=-5
Find the zeros of 15=7x+2x^2
The zeros of 15=7x+2x^2 are x= 3/2 ,x=-5