What are the solutions to the equation 2x^2+2x-312=0 ?

The solutions to the equation 2x^2+2x-312=0 are x=12,x=-13

Find the zeros of 2x^2+2x-312=0

The zeros of 2x^2+2x-312=0 are x=12,x=-13

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2 x^{2} + 2 x - 312 = 0

x = 12, x = - 13

Solution steps

2 x^{2} + 2 x - 312 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 2 ( - 312 )}{2 \cdot 2}

2^{2} - 4 \cdot 2 (- 312) = 50

x_{1, 2} = \frac{- 2 \pm 50}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 2 + 50}{2 \cdot 2}, x_{2} = \frac{- 2 - 50}{2 \cdot 2}

x = \frac{- 2 + 50}{2 \cdot 2} : 12

x = \frac{- 2 - 50}{2 \cdot 2} : - 13

The solutions to the quadratic equation are:

x = 12, x = - 13

t^{2} - 13 t + 36 = 0

q, - 0.3 q^{2} + 36 q = - 3 q - 100

\frac{1}{8} x^{2} + 2 x + 16 = 0

2 (x + 1)^{2} = 18

x, 3 x^{2} - 6 y^{2} + 3 x y + x + 2 y = 0

What are the solutions to the equation 2x^2+2x-312=0 ?
The solutions to the equation 2x^2+2x-312=0 are x=12,x=-13
Find the zeros of 2x^2+2x-312=0
The zeros of 2x^2+2x-312=0 are x=12,x=-13