What are the solutions to the equation 3x^2+8x-7=4 ?

The solutions to the equation 3x^2+8x-7=4 are x=1,x=-11/3

Find the zeros of 3x^2+8x-7=4

The zeros of 3x^2+8x-7=4 are x=1,x=-11/3

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3 x^{2} + 8 x - 7 = 4

x = 1, x = - \frac{11}{3}

Solution steps

3 x^{2} + 8 x - 7 = 4

Move

4

to the left side

3 x^{2} + 8 x - 11 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 3 ( - 11 )}{2 \cdot 3}

8^{2} - 4 \cdot 3 (- 11) = 14

x_{1, 2} = \frac{- 8 \pm 14}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 8 + 14}{2 \cdot 3}, x_{2} = \frac{- 8 - 14}{2 \cdot 3}

x = \frac{- 8 + 14}{2 \cdot 3} : 1

x = \frac{- 8 - 14}{2 \cdot 3} : - \frac{11}{3}

The solutions to the quadratic equation are:

x = 1, x = - \frac{11}{3}

x, y + 3 x^{2} y = x^{2}

x, x^{2} - 4 x - 5 = 0

5 v^{2} + 3 = 3 v

(x - 6) (x - 5) = 0

- x^{2} + 4 x - 3 = 0

What are the solutions to the equation 3x^2+8x-7=4 ?
The solutions to the equation 3x^2+8x-7=4 are x=1,x=-11/3
Find the zeros of 3x^2+8x-7=4
The zeros of 3x^2+8x-7=4 are x=1,x=-11/3