What are the solutions to the equation 2x^2+16x-20=0 ?

The solutions to the equation 2x^2+16x-20=0 are x=-4+sqrt(26),x=-4-sqrt(26)

Find the zeros of 2x^2+16x-20=0

The zeros of 2x^2+16x-20=0 are x=-4+sqrt(26),x=-4-sqrt(26)

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2 x^{2} + 16 x - 20 = 0

x = - 4 + 26, x = - 4 - 26

Decimal Notation

x = 1.09901 \dots, x = - 9.09901 \dots

Solution steps

2 x^{2} + 16 x - 20 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 \cdot 2 ( - 20 )}{2 \cdot 2}

1 6^{2} - 4 \cdot 2 (- 20) = 426

x_{1, 2} = \frac{- 16 \pm 4 26}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 16 + 4 26}{2 \cdot 2}, x_{2} = \frac{- 16 - 4 26}{2 \cdot 2}

x = \frac{- 16 + 4 26}{2 \cdot 2} : - 4 + 26

x = \frac{- 16 - 4 26}{2 \cdot 2} : - 4 - 26

The solutions to the quadratic equation are:

x = - 4 + 26, x = - 4 - 26

x, 4 x^{2} + 2 x y + y^{2} = 0

2 w^{2} + 11 w - 63 = 0

1 + 0.2 x - 0.48 x^{2} = 0

x, \frac{1}{2} + \frac{x ^{2}}{32} = R

0 = - 0.3 x^{2} + 22 x + 20

What are the solutions to the equation 2x^2+16x-20=0 ?
The solutions to the equation 2x^2+16x-20=0 are x=-4+sqrt(26),x=-4-sqrt(26)
Find the zeros of 2x^2+16x-20=0
The zeros of 2x^2+16x-20=0 are x=-4+sqrt(26),x=-4-sqrt(26)