What are the solutions to the equation 0=x^2+33x-280 ?

The solutions to the equation 0=x^2+33x-280 are x=7,x=-40

Find the zeros of 0=x^2+33x-280

The zeros of 0=x^2+33x-280 are x=7,x=-40

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0 = x^{2} + 33 x - 280

x = 7, x = - 40

Solution steps

0 = x^{2} + 33 x - 280

Switch sides

x^{2} + 33 x - 280 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 33 \pm 3 3 ^{2} - 4 \cdot 1 \cdot ( - 280 )}{2 \cdot 1}

3 3^{2} - 4 \cdot 1 \cdot (- 280) = 47

x_{1, 2} = \frac{- 33 \pm 47}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 33 + 47}{2 \cdot 1}, x_{2} = \frac{- 33 - 47}{2 \cdot 1}

x = \frac{- 33 + 47}{2 \cdot 1} : 7

x = \frac{- 33 - 47}{2 \cdot 1} : - 40

The solutions to the quadratic equation are:

x = 7, x = - 40

1 - 0.4 L - 0.2 L^{2} = 0

x^{2} - 13 + 2 x + 49 = 116

x^{2} + 3 x + 2.25 = 0

3 x^{2} - 11 x = - 7 x - 4

10 x^{2} - 19 x - 12 = 0

What are the solutions to the equation 0=x^2+33x-280 ?
The solutions to the equation 0=x^2+33x-280 are x=7,x=-40
Find the zeros of 0=x^2+33x-280
The zeros of 0=x^2+33x-280 are x=7,x=-40