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

The solutions to the equation (x)=2x^2+4x-3 are x=(-3+sqrt(33))/4 ,x=(-3-sqrt(33))/4

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

The zeros of (x)=2x^2+4x-3 are x=(-3+sqrt(33))/4 ,x=(-3-sqrt(33))/4

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(x) = 2 x^{2} + 4 x - 3

x = \frac{- 3 + 33}{4}, x = \frac{- 3 - 33}{4}

Decimal Notation

x = 0.68614 \dots, x = - 2.18614 \dots

Solution steps

(x) = 2 x^{2} + 4 x - 3

Refine

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

Switch sides

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

Move

x

to the left side

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

Solve with the quadratic formula

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

3^{2} - 4 \cdot 2 (- 3) = 33

x_{1, 2} = \frac{- 3 \pm 33}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 3 + 33}{2 \cdot 2}, x_{2} = \frac{- 3 - 33}{2 \cdot 2}

x = \frac{- 3 + 33}{2 \cdot 2} : \frac{- 3 + 33}{4}

x = \frac{- 3 - 33}{2 \cdot 2} : \frac{- 3 - 33}{4}

The solutions to the quadratic equation are:

x = \frac{- 3 + 33}{4}, x = \frac{- 3 - 33}{4}

x, 4 x^{2} + 4 y^{2} - 6 x + 8 y + 21 = 0

x^{2} + (3 m - 15) x + 12 = 0

3 x^{2} + 7 x = 10

y^{2} + 10 y + 21 = 0

- a^{2} - a + 6 = 0

What are the solutions to the equation (x)=2x^2+4x-3 ?
The solutions to the equation (x)=2x^2+4x-3 are x=(-3+sqrt(33))/4 ,x=(-3-sqrt(33))/4
Find the zeros of (x)=2x^2+4x-3
The zeros of (x)=2x^2+4x-3 are x=(-3+sqrt(33))/4 ,x=(-3-sqrt(33))/4