What are the solutions to the equation x^2+4x+c=21 ?

The solutions to the equation x^2+4x+c=21 are x=-2+sqrt(-c+25),x=-sqrt(-c+25)-2

Find the zeros of x^2+4x+c=21

The zeros of x^2+4x+c=21 are x=-2+sqrt(-c+25),x=-sqrt(-c+25)-2

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x^{2} + 4 x + c = 21

x = - 2 + - c + 25, x = - - c + 25 - 2

Solution steps

x^{2} + 4 x + c = 21

Move

21

to the left side

x^{2} + 4 x + c - 21 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( c - 21 )}{2 \cdot 1}

Simplify

4^{2} - 4 \cdot 1 \cdot (c - 21) : 2 - c + 25

x_{1, 2} = \frac{- 4 \pm 2 - c + 25}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 4 + 2 - c + 25}{2 \cdot 1}, x_{2} = \frac{- 4 - 2 - c + 25}{2 \cdot 1}

x = \frac{- 4 + 2 - c + 25}{2 \cdot 1} : - 2 + - c + 25

x = \frac{- 4 - 2 - c + 25}{2 \cdot 1} : - - c + 25 - 2

The solutions to the quadratic equation are:

x = - 2 + - c + 25, x = - - c + 25 - 2

9 x^{2} - 9 x + 4 = 0

x^{2} - 3 a x - 28 a^{2} = 0

2 x^{2} - 3 x = 7

y, 3 x^{2} - y^{2} = 5 x^{3} y^{2} - 4

x, 3 x^{2} - 8 x + 4 = 0

What are the solutions to the equation x^2+4x+c=21 ?
The solutions to the equation x^2+4x+c=21 are x=-2+sqrt(-c+25),x=-sqrt(-c+25)-2
Find the zeros of x^2+4x+c=21
The zeros of x^2+4x+c=21 are x=-2+sqrt(-c+25),x=-sqrt(-c+25)-2