What are the solutions to the equation (2x+3)x=4 ?

The solutions to the equation (2x+3)x=4 are x=(-3+sqrt(41))/4 ,x=(-3-sqrt(41))/4

Find the zeros of (2x+3)x=4

The zeros of (2x+3)x=4 are x=(-3+sqrt(41))/4 ,x=(-3-sqrt(41))/4

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(2 x + 3) x = 4

x = \frac{- 3 + 41}{4}, x = \frac{- 3 - 41}{4}

Decimal Notation

x = 0.85078 \dots, x = - 2.35078 \dots

Solution steps

(2 x + 3) x = 4

Expand

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

2 x^{2} + 3 x = 4

Move

4

to the left side

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

Solve with the quadratic formula

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

3^{2} - 4 \cdot 2 (- 4) = 41

x_{1, 2} = \frac{- 3 \pm 41}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 3 + 41}{2 \cdot 2}, x_{2} = \frac{- 3 - 41}{2 \cdot 2}

x = \frac{- 3 + 41}{2 \cdot 2} : \frac{- 3 + 41}{4}

x = \frac{- 3 - 41}{2 \cdot 2} : \frac{- 3 - 41}{4}

The solutions to the quadratic equation are:

x = \frac{- 3 + 41}{4}, x = \frac{- 3 - 41}{4}

x^{2} - 60 = 4 x

9 (2 - x) = 2 x^{2}

5 m^{2} + 1 = 0

x^{2} - 10 x + 40 = 0

104 x^{2} - 75 x - 14 = 0

What are the solutions to the equation (2x+3)x=4 ?
The solutions to the equation (2x+3)x=4 are x=(-3+sqrt(41))/4 ,x=(-3-sqrt(41))/4
Find the zeros of (2x+3)x=4
The zeros of (2x+3)x=4 are x=(-3+sqrt(41))/4 ,x=(-3-sqrt(41))/4