What are the solutions to the equation 4x^2+20x+25=49 ?

The solutions to the equation 4x^2+20x+25=49 are x=1,x=-6

Find the zeros of 4x^2+20x+25=49

The zeros of 4x^2+20x+25=49 are x=1,x=-6

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4 x^{2} + 20 x + 25 = 49

x = 1, x = - 6

Solution steps

4 x^{2} + 20 x + 25 = 49

Move

49

to the left side

4 x^{2} + 20 x - 24 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 4 ( - 24 )}{2 \cdot 4}

2 0^{2} - 4 \cdot 4 (- 24) = 28

x_{1, 2} = \frac{- 20 \pm 28}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 20 + 28}{2 \cdot 4}, x_{2} = \frac{- 20 - 28}{2 \cdot 4}

x = \frac{- 20 + 28}{2 \cdot 4} : 1

x = \frac{- 20 - 28}{2 \cdot 4} : - 6

The solutions to the quadratic equation are:

x = 1, x = - 6

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

x^{2} + 4 x - 21 \geq 0

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

(- 6 a^{2} b) \cdot (\frac{1}{2} ab)

\frac{1}{2} \cdot 72 \cdot (0.05 - 0.03)

What are the solutions to the equation 4x^2+20x+25=49 ?
The solutions to the equation 4x^2+20x+25=49 are x=1,x=-6
Find the zeros of 4x^2+20x+25=49
The zeros of 4x^2+20x+25=49 are x=1,x=-6