What are the solutions to the equation 4x^2+12x+9=49 ?

The solutions to the equation 4x^2+12x+9=49 are x=2,x=-5

Find the zeros of 4x^2+12x+9=49

The zeros of 4x^2+12x+9=49 are x=2,x=-5

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4 x^{2} + 12 x + 9 = 49

x = 2, x = - 5

Solution steps

4 x^{2} + 12 x + 9 = 49

Move

49

to the left side

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

Solve with the quadratic formula

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

1 2^{2} - 4 \cdot 4 (- 40) = 28

x_{1, 2} = \frac{- 12 \pm 28}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 12 + 28}{2 \cdot 4}, x_{2} = \frac{- 12 - 28}{2 \cdot 4}

x = \frac{- 12 + 28}{2 \cdot 4} : 2

x = \frac{- 12 - 28}{2 \cdot 4} : - 5

The solutions to the quadratic equation are:

x = 2, x = - 5

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

x, x^{2} + y^{2} - 4 x - 2 y + 36 = 0

2 p^{2} - 7 p + 3 = 0

9 x^{2} = 9

0 = - 16 x^{2} - 32 x - 20

What are the solutions to the equation 4x^2+12x+9=49 ?
The solutions to the equation 4x^2+12x+9=49 are x=2,x=-5
Find the zeros of 4x^2+12x+9=49
The zeros of 4x^2+12x+9=49 are x=2,x=-5