What are the solutions to the equation 2x^2+22x+60=0 ?

The solutions to the equation 2x^2+22x+60=0 are x=-5,x=-6

Find the zeros of 2x^2+22x+60=0

The zeros of 2x^2+22x+60=0 are x=-5,x=-6

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2 x^{2} + 22 x + 60 = 0

x = - 5, x = - 6

Solution steps

2 x^{2} + 22 x + 60 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 22 \pm 2 2 ^{2} - 4 \cdot 2 \cdot 60}{2 \cdot 2}

2 2^{2} - 4 \cdot 2 \cdot 60 = 2

x_{1, 2} = \frac{- 22 \pm 2}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 22 + 2}{2 \cdot 2}, x_{2} = \frac{- 22 - 2}{2 \cdot 2}

x = \frac{- 22 + 2}{2 \cdot 2} : - 5

x = \frac{- 22 - 2}{2 \cdot 2} : - 6

The solutions to the quadratic equation are:

x = - 5, x = - 6

(\frac{5}{6})^{6}

x^{2} = 29

e^{- 0.2496}

10000 0^{\frac{2}{5}}

195 c^{6} p^{5} - 91 c^{5} p^{6} + 221 c^{3} p^{10}

What are the solutions to the equation 2x^2+22x+60=0 ?
The solutions to the equation 2x^2+22x+60=0 are x=-5,x=-6
Find the zeros of 2x^2+22x+60=0
The zeros of 2x^2+22x+60=0 are x=-5,x=-6