What are the solutions to the equation 0=3x^2+12x-15 ?

The solutions to the equation 0=3x^2+12x-15 are x=1,x=-5

Find the zeros of 0=3x^2+12x-15

The zeros of 0=3x^2+12x-15 are x=1,x=-5

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

x = 1, x = - 5

Solution steps

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

Switch sides

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 3 ( - 15 )}{2 \cdot 3}

1 2^{2} - 4 \cdot 3 (- 15) = 18

x_{1, 2} = \frac{- 12 \pm 18}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 12 + 18}{2 \cdot 3}, x_{2} = \frac{- 12 - 18}{2 \cdot 3}

x = \frac{- 12 + 18}{2 \cdot 3} : 1

x = \frac{- 12 - 18}{2 \cdot 3} : - 5

The solutions to the quadratic equation are:

x = 1, x = - 5

(2 x + 1) (x - 3) = - 4 x

8 x^{2} - 2 x - 10 = 0

x^{2} + 6 - 8 = 0

3 3^{2} + 5 6^{2} = x^{2}

x^{2} = 3 - x

What are the solutions to the equation 0=3x^2+12x-15 ?
The solutions to the equation 0=3x^2+12x-15 are x=1,x=-5
Find the zeros of 0=3x^2+12x-15
The zeros of 0=3x^2+12x-15 are x=1,x=-5