What are the solutions to the equation x^2+13x+40=4 ?

The solutions to the equation x^2+13x+40=4 are x=-4,x=-9

Find the zeros of x^2+13x+40=4

The zeros of x^2+13x+40=4 are x=-4,x=-9

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x^{2} + 13 x + 40 = 4

x = - 4, x = - 9

Solution steps

x^{2} + 13 x + 40 = 4

Move

4

to the left side

x^{2} + 13 x + 36 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 13 \pm 1 3 ^{2} - 4 \cdot 1 \cdot 36}{2 \cdot 1}

1 3^{2} - 4 \cdot 1 \cdot 36 = 5

x_{1, 2} = \frac{- 13 \pm 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 13 + 5}{2 \cdot 1}, x_{2} = \frac{- 13 - 5}{2 \cdot 1}

x = \frac{- 13 + 5}{2 \cdot 1} : - 4

x = \frac{- 13 - 5}{2 \cdot 1} : - 9

The solutions to the quadratic equation are:

x = - 4, x = - 9

(x - 8) (x - 1) = 0

x^{2} + 2 x - 5 = 0

0 = (x - 3)^{2} + 1

x, z = 4 - x^{2} - 2 y^{2}

x, x^{2} + m x + 7 = 0

What are the solutions to the equation x^2+13x+40=4 ?
The solutions to the equation x^2+13x+40=4 are x=-4,x=-9
Find the zeros of x^2+13x+40=4
The zeros of x^2+13x+40=4 are x=-4,x=-9