What are the solutions to the equation 4x^2+4x+1=36 ?

The solutions to the equation 4x^2+4x+1=36 are x= 5/2 ,x=-7/2

Find the zeros of 4x^2+4x+1=36

The zeros of 4x^2+4x+1=36 are x= 5/2 ,x=-7/2

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4 x^{2} + 4 x + 1 = 36

x = \frac{5}{2}, x = - \frac{7}{2}

Decimal Notation

x = 2.5, x = - 3.5

Solution steps

4 x^{2} + 4 x + 1 = 36

Move

36

to the left side

4 x^{2} + 4 x - 35 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 4 ( - 35 )}{2 \cdot 4}

4^{2} - 4 \cdot 4 (- 35) = 24

x_{1, 2} = \frac{- 4 \pm 24}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 4 + 24}{2 \cdot 4}, x_{2} = \frac{- 4 - 24}{2 \cdot 4}

x = \frac{- 4 + 24}{2 \cdot 4} : \frac{5}{2}

x = \frac{- 4 - 24}{2 \cdot 4} : - \frac{7}{2}

The solutions to the quadratic equation are:

x = \frac{5}{2}, x = - \frac{7}{2}

x, (x + y)^{2} = 64

4 - 11 x = 3 x^{2}

x, x^{2} + 9 y^{2} = 0

x, (x^{2} + y^{2}) (p^{2} + q^{2}) = 1

a, a^{2} \cdot b^{2} = c^{2}

What are the solutions to the equation 4x^2+4x+1=36 ?
The solutions to the equation 4x^2+4x+1=36 are x= 5/2 ,x=-7/2
Find the zeros of 4x^2+4x+1=36
The zeros of 4x^2+4x+1=36 are x= 5/2 ,x=-7/2