What are the solutions to the equation 7x^2+40x=12 ?

The solutions to the equation 7x^2+40x=12 are x= 2/7 ,x=-6

Find the zeros of 7x^2+40x=12

The zeros of 7x^2+40x=12 are x= 2/7 ,x=-6

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7 x^{2} + 40 x = 12

x = \frac{2}{7}, x = - 6

Decimal Notation

x = 0.28571 \dots, x = - 6

Solution steps

7 x^{2} + 40 x = 12

Move

12

to the left side

7 x^{2} + 40 x - 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 40 \pm 4 0 ^{2} - 4 \cdot 7 ( - 12 )}{2 \cdot 7}

4 0^{2} - 4 \cdot 7 (- 12) = 44

x_{1, 2} = \frac{- 40 \pm 44}{2 \cdot 7}

Separate the solutions

x_{1} = \frac{- 40 + 44}{2 \cdot 7}, x_{2} = \frac{- 40 - 44}{2 \cdot 7}

x = \frac{- 40 + 44}{2 \cdot 7} : \frac{2}{7}

x = \frac{- 40 - 44}{2 \cdot 7} : - 6

The solutions to the quadratic equation are:

x = \frac{2}{7}, x = - 6

(2 x - 5)^{2} - (x - 6)^{2} = 5

3 x^{2} - 6 x + 14

2 x^{2} - 36 x = - 160

(h - 2)^{2} = 9

\frac{1}{5} x^{2} + 14 = 59

What are the solutions to the equation 7x^2+40x=12 ?
The solutions to the equation 7x^2+40x=12 are x= 2/7 ,x=-6
Find the zeros of 7x^2+40x=12
The zeros of 7x^2+40x=12 are x= 2/7 ,x=-6