What are the solutions to the equation x^2+18x+96=3x+40 ?

The solutions to the equation x^2+18x+96=3x+40 are x=-7,x=-8

Find the zeros of x^2+18x+96=3x+40

The zeros of x^2+18x+96=3x+40 are x=-7,x=-8

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x^{2} + 18 x + 96 = 3 x + 40

x = - 7, x = - 8

Solution steps

x^{2} + 18 x + 96 = 3 x + 40

Move

40

to the left side

x^{2} + 18 x + 56 = 3 x

Move

3 x

to the left side

x^{2} + 15 x + 56 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 15 \pm 1 5 ^{2} - 4 \cdot 1 \cdot 56}{2 \cdot 1}

1 5^{2} - 4 \cdot 1 \cdot 56 = 1

x_{1, 2} = \frac{- 15 \pm 1}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 15 + 1}{2 \cdot 1}, x_{2} = \frac{- 15 - 1}{2 \cdot 1}

x = \frac{- 15 + 1}{2 \cdot 1} : - 7

x = \frac{- 15 - 1}{2 \cdot 1} : - 8

The solutions to the quadratic equation are:

x = - 7, x = - 8

∣ 8 x + 4 ∣ < 7

4 y + 3 > 2 y + 14

- 4 + 8 a - 4 a^{2}

10 n^{3} - 9 n^{2} + n

- 3 x^{2} + 5 x - 7 = 0

What are the solutions to the equation x^2+18x+96=3x+40 ?
The solutions to the equation x^2+18x+96=3x+40 are x=-7,x=-8
Find the zeros of x^2+18x+96=3x+40
The zeros of x^2+18x+96=3x+40 are x=-7,x=-8