What are the solutions to the equation 2x^2+3x-3=1 ?

The solutions to the equation 2x^2+3x-3=1 are x=(-3+sqrt(41))/4 ,x=(-3-sqrt(41))/4

Find the zeros of 2x^2+3x-3=1

The zeros of 2x^2+3x-3=1 are x=(-3+sqrt(41))/4 ,x=(-3-sqrt(41))/4

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2 x^{2} + 3 x - 3 = 1

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

Decimal Notation

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

Solution steps

2 x^{2} + 3 x - 3 = 1

Move

1

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}

5 x^{2} - 9 x - 10 = 0

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

9 x^{2} + 5 = 1

x^{2} + 6 x - 150 = 0

v^{2} - 12 v + 35 = 0

What are the solutions to the equation 2x^2+3x-3=1 ?
The solutions to the equation 2x^2+3x-3=1 are x=(-3+sqrt(41))/4 ,x=(-3-sqrt(41))/4
Find the zeros of 2x^2+3x-3=1
The zeros of 2x^2+3x-3=1 are x=(-3+sqrt(41))/4 ,x=(-3-sqrt(41))/4