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

The solutions to the equation 5=x(2x+3) are x=1,x=-5/2

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

The zeros of 5=x(2x+3) are x=1,x=-5/2

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5 = x (2 x + 3)

x = 1, x = - \frac{5}{2}

Decimal Notation

x = 1, x = - 2.5

Solution steps

5 = x (2 x + 3)

Expand

x (2 x + 3) : 2 x^{2} + 3 x

5 = 2 x^{2} + 3 x

Switch sides

2 x^{2} + 3 x = 5

Move

5

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 2 ( - 5 )}{2 \cdot 2}

3^{2} - 4 \cdot 2 (- 5) = 7

x_{1, 2} = \frac{- 3 \pm 7}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 3 + 7}{2 \cdot 2}, x_{2} = \frac{- 3 - 7}{2 \cdot 2}

x = \frac{- 3 + 7}{2 \cdot 2} : 1

x = \frac{- 3 - 7}{2 \cdot 2} : - \frac{5}{2}

The solutions to the quadratic equation are:

x = 1, x = - \frac{5}{2}

13 + 2 v - 8 + 6 > - 7 - v

x + 8 = 21

3 x^{2} - 6 x < 2 x^{2} + 7

x (x + 10) = 0

\frac{3 - x}{x - 4} \geq - 4

What are the solutions to the equation 5=x(2x+3) ?
The solutions to the equation 5=x(2x+3) are x=1,x=-5/2
Find the zeros of 5=x(2x+3)
The zeros of 5=x(2x+3) are x=1,x=-5/2