What are the solutions to the equation x^2+4x+21=42 ?

The solutions to the equation x^2+4x+21=42 are x=3,x=-7

Find the zeros of x^2+4x+21=42

The zeros of x^2+4x+21=42 are x=3,x=-7

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x^{2} + 4 x + 21 = 42

x = 3, x = - 7

Solution steps

x^{2} + 4 x + 21 = 42

Move

42

to the left side

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

Solve with the quadratic formula

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

4^{2} - 4 \cdot 1 \cdot (- 21) = 10

x_{1, 2} = \frac{- 4 \pm 10}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 4 + 10}{2 \cdot 1}, x_{2} = \frac{- 4 - 10}{2 \cdot 1}

x = \frac{- 4 + 10}{2 \cdot 1} : 3

x = \frac{- 4 - 10}{2 \cdot 1} : - 7

The solutions to the quadratic equation are:

x = 3, x = - 7

9 (3 x + 9) + (x + 3) = 14

\frac{\frac{2 x + y}{36}}{\frac{36 + 36 x - 7 x ^{2}}{288}}

- \frac{11}{2 ( - 3 + 4 ) ^{2}}

392

x^{2} < x + 42

What are the solutions to the equation x^2+4x+21=42 ?
The solutions to the equation x^2+4x+21=42 are x=3,x=-7
Find the zeros of x^2+4x+21=42
The zeros of x^2+4x+21=42 are x=3,x=-7