What are the solutions to the equation 2x^2+12x=3 ?

The solutions to the equation 2x^2+12x=3 are x=(-6+sqrt(42))/2 ,x=-(6+sqrt(42))/2

Find the zeros of 2x^2+12x=3

The zeros of 2x^2+12x=3 are x=(-6+sqrt(42))/2 ,x=-(6+sqrt(42))/2

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2 x^{2} + 12 x = 3

x = \frac{- 6 + 42}{2}, x = - \frac{6 + 42}{2}

Decimal Notation

x = 0.24037 \dots, x = - 6.24037 \dots

Solution steps

2 x^{2} + 12 x = 3

Move

3

to the left side

2 x^{2} + 12 x - 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 2 ( - 3 )}{2 \cdot 2}

1 2^{2} - 4 \cdot 2 (- 3) = 242

x_{1, 2} = \frac{- 12 \pm 2 42}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 12 + 2 42}{2 \cdot 2}, x_{2} = \frac{- 12 - 2 42}{2 \cdot 2}

x = \frac{- 12 + 2 42}{2 \cdot 2} : \frac{- 6 + 42}{2}

x = \frac{- 12 - 2 42}{2 \cdot 2} : - \frac{6 + 42}{2}

The solutions to the quadratic equation are:

x = \frac{- 6 + 42}{2}, x = - \frac{6 + 42}{2}

r, A = 4 r^{2}

y^{2} - 8 y - 4 = 0

b^{2} - 10 b + 21 = 0

x (2 x - 3) = 18

- 5 x^{2} + 60 x - 405 = 0

What are the solutions to the equation 2x^2+12x=3 ?
The solutions to the equation 2x^2+12x=3 are x=(-6+sqrt(42))/2 ,x=-(6+sqrt(42))/2
Find the zeros of 2x^2+12x=3
The zeros of 2x^2+12x=3 are x=(-6+sqrt(42))/2 ,x=-(6+sqrt(42))/2