What are the solutions to the equation 4x^2+5x+3=2x^2-3x ?

The solutions to the equation 4x^2+5x+3=2x^2-3x are x=(-4+sqrt(10))/2 ,x=-(4+sqrt(10))/2

Find the zeros of 4x^2+5x+3=2x^2-3x

The zeros of 4x^2+5x+3=2x^2-3x are x=(-4+sqrt(10))/2 ,x=-(4+sqrt(10))/2

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

x = \frac{- 4 + 10}{2}, x = - \frac{4 + 10}{2}

Decimal

x = - 0.41886 \dots, x = - 3.58113 \dots

Solution steps

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

Move

3 x

to the left side

4 x^{2} + 8 x + 3 = 2 x^{2}

Move

2 x^{2}

to the left side

2 x^{2} + 8 x + 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 2 \cdot 3}{2 \cdot 2}

8^{2} - 4 \cdot 2 \cdot 3 = 210

x_{1, 2} = \frac{- 8 \pm 2 10}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 8 + 2 10}{2 \cdot 2}, x_{2} = \frac{- 8 - 2 10}{2 \cdot 2}

x = \frac{- 8 + 2 10}{2 \cdot 2} : \frac{- 4 + 10}{2}

x = \frac{- 8 - 2 10}{2 \cdot 2} : - \frac{4 + 10}{2}

The solutions to the quadratic equation are:

x = \frac{- 4 + 10}{2}, x = - \frac{4 + 10}{2}

\frac{2}{2} \cdot \frac{1}{2}

∣ x + 2 ∣ - x \geq 0

x^{2} + 36 = 0

(\frac{a}{b})^{- n}

x^{2} - 12 x - 14 = 0

What are the solutions to the equation 4x^2+5x+3=2x^2-3x ?
The solutions to the equation 4x^2+5x+3=2x^2-3x are x=(-4+sqrt(10))/2 ,x=-(4+sqrt(10))/2
Find the zeros of 4x^2+5x+3=2x^2-3x
The zeros of 4x^2+5x+3=2x^2-3x are x=(-4+sqrt(10))/2 ,x=-(4+sqrt(10))/2