What are the solutions to the equation 42=(x+6)(x+4) ?

The solutions to the equation 42=(x+6)(x+4) are x=-5+sqrt(43),x=-5-sqrt(43)

Find the zeros of 42=(x+6)(x+4)

The zeros of 42=(x+6)(x+4) are x=-5+sqrt(43),x=-5-sqrt(43)

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42 = (x + 6) (x + 4)

x = - 5 + 43, x = - 5 - 43

Decimal Notation

x = 1.55743 \dots, x = - 11.55743 \dots

Solution steps

42 = (x + 6) (x + 4)

Expand

(x + 6) (x + 4) : x^{2} + 10 x + 24

42 = x^{2} + 10 x + 24

Switch sides

x^{2} + 10 x + 24 = 42

Move

42

to the left side

x^{2} + 10 x - 18 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 1 \cdot ( - 18 )}{2 \cdot 1}

1 0^{2} - 4 \cdot 1 \cdot (- 18) = 243

x_{1, 2} = \frac{- 10 \pm 2 43}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 10 + 2 43}{2 \cdot 1}, x_{2} = \frac{- 10 - 2 43}{2 \cdot 1}

x = \frac{- 10 + 2 43}{2 \cdot 1} : - 5 + 43

x = \frac{- 10 - 2 43}{2 \cdot 1} : - 5 - 43

The solutions to the quadratic equation are:

x = - 5 + 43, x = - 5 - 43

300 0^{2} + 400 0^{2} = x^{2}

2 x^{2} + 4 x + 96 = 0

x, 64 x^{2} - 48 x y + 9 y^{2} = 0

x^{2} (- 2) = 16

x^{2} + x + 32 = 0

What are the solutions to the equation 42=(x+6)(x+4) ?
The solutions to the equation 42=(x+6)(x+4) are x=-5+sqrt(43),x=-5-sqrt(43)
Find the zeros of 42=(x+6)(x+4)
The zeros of 42=(x+6)(x+4) are x=-5+sqrt(43),x=-5-sqrt(43)