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

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

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

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

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

x = - 5, x = 6

Solution steps

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

Solve with the quadratic formula

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

2^{2} - 4 (- 2) \cdot 60 = 22

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

Separate the solutions

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

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

x = \frac{- 2 - 22}{2 ( - 2 )} : 6

The solutions to the quadratic equation are:

x = - 5, x = 6

y^{- 3}

p + 4 - 13 = - 6

\frac{4 x y ^{2}}{16 x y}

∣ - 12 ∣

8 x^{3} - y^{6}

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