What are the solutions to the equation 3x^2+12x-5=7 ?

The solutions to the equation 3x^2+12x-5=7 are x=2(sqrt(2)-1),x=-2(1+sqrt(2))

Find the zeros of 3x^2+12x-5=7

The zeros of 3x^2+12x-5=7 are x=2(sqrt(2)-1),x=-2(1+sqrt(2))

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3 x^{2} + 12 x - 5 = 7

x = 2 (2 - 1), x = - 2 (1 + 2)

Decimal Notation

x = 0.82842 \dots, x = - 4.82842 \dots

Solution steps

3 x^{2} + 12 x - 5 = 7

Move

7

to the left side

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

Solve with the quadratic formula

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

1 2^{2} - 4 \cdot 3 (- 12) = 122

x_{1, 2} = \frac{- 12 \pm 12 2}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 12 + 12 2}{2 \cdot 3}, x_{2} = \frac{- 12 - 12 2}{2 \cdot 3}

x = \frac{- 12 + 12 2}{2 \cdot 3} : 2 (2 - 1)

x = \frac{- 12 - 12 2}{2 \cdot 3} : - 2 (1 + 2)

The solutions to the quadratic equation are:

x = 2 (2 - 1), x = - 2 (1 + 2)

\frac{3}{4} \cdot \frac{2}{9}

\frac{( x + h ) ^{2} - 3 ( x + h ) - ( x ^{2} - 3 x )}{h}

- \frac{40}{2}

24 m - 12 p + 72

x^{- 1} \cdot x

What are the solutions to the equation 3x^2+12x-5=7 ?
The solutions to the equation 3x^2+12x-5=7 are x=2(sqrt(2)-1),x=-2(1+sqrt(2))
Find the zeros of 3x^2+12x-5=7
The zeros of 3x^2+12x-5=7 are x=2(sqrt(2)-1),x=-2(1+sqrt(2))