What are the solutions to the equation 4x^2+7x+1=3 ?

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

Find the zeros of 4x^2+7x+1=3

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

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4 x^{2} + 7 x + 1 = 3

x = \frac{1}{4}, x = - 2

Solution steps

4 x^{2} + 7 x + 1 = 3

Move

3

to the left side

4 x^{2} + 7 x - 2 = 0

Solve with the quadratic formula

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

7^{2} - 4 \cdot 4 (- 2) = 9

x_{1, 2} = \frac{- 7 \pm 9}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 7 + 9}{2 \cdot 4}, x_{2} = \frac{- 7 - 9}{2 \cdot 4}

x = \frac{- 7 + 9}{2 \cdot 4} : \frac{1}{4}

x = \frac{- 7 - 9}{2 \cdot 4} : - 2

The solutions to the quadratic equation are:

x = \frac{1}{4}, x = - 2

- 48 x^{2} - 6 = 0

(x + 5) (4 x - 3) = (x + 3) (2 x + 10)

x^{2} + 9 x + 7 = 0

4 y^{2} = 196

- - 40

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