What are the solutions to the equation 2x^2+20x-52=-4 ?

The solutions to the equation 2x^2+20x-52=-4 are x=2,x=-12

Find the zeros of 2x^2+20x-52=-4

The zeros of 2x^2+20x-52=-4 are x=2,x=-12

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2 x^{2} + 20 x - 52 = - 4

x = 2, x = - 12

Solution steps

2 x^{2} + 20 x - 52 = - 4

Move

4

to the left side

2 x^{2} + 20 x - 48 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 2 ( - 48 )}{2 \cdot 2}

2 0^{2} - 4 \cdot 2 (- 48) = 28

x_{1, 2} = \frac{- 20 \pm 28}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 20 + 28}{2 \cdot 2}, x_{2} = \frac{- 20 - 28}{2 \cdot 2}

x = \frac{- 20 + 28}{2 \cdot 2} : 2

x = \frac{- 20 - 28}{2 \cdot 2} : - 12

The solutions to the quadratic equation are:

x = 2, x = - 12

x^{2} \leq 8 x - 11

- 6 x - 7 = 5 x - 117

30 x^{2} - 50 x

49 + 9

(9 m n^{6})^{\frac{1}{2}}

What are the solutions to the equation 2x^2+20x-52=-4 ?
The solutions to the equation 2x^2+20x-52=-4 are x=2,x=-12
Find the zeros of 2x^2+20x-52=-4
The zeros of 2x^2+20x-52=-4 are x=2,x=-12