What are the solutions to the equation 2x^2+7x-555=0 ?

The solutions to the equation 2x^2+7x-555=0 are x=15,x=-37/2

Find the zeros of 2x^2+7x-555=0

The zeros of 2x^2+7x-555=0 are x=15,x=-37/2

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2 x^{2} + 7 x - 555 = 0

x = 15, x = - \frac{37}{2}

Decimal Notation

x = 15, x = - 18.5

Solution steps

2 x^{2} + 7 x - 555 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 2 ( - 555 )}{2 \cdot 2}

7^{2} - 4 \cdot 2 (- 555) = 67

x_{1, 2} = \frac{- 7 \pm 67}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 7 + 67}{2 \cdot 2}, x_{2} = \frac{- 7 - 67}{2 \cdot 2}

x = \frac{- 7 + 67}{2 \cdot 2} : 15

x = \frac{- 7 - 67}{2 \cdot 2} : - \frac{37}{2}

The solutions to the quadratic equation are:

x = 15, x = - \frac{37}{2}

t, S = u t + \frac{1}{2} a t^{2}

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

x^{2} + 3 x - 8 = 0

12 x^{2} + 19 x = 21

775 = - q^{2} + 1000

What are the solutions to the equation 2x^2+7x-555=0 ?
The solutions to the equation 2x^2+7x-555=0 are x=15,x=-37/2
Find the zeros of 2x^2+7x-555=0
The zeros of 2x^2+7x-555=0 are x=15,x=-37/2