What are the solutions to the equation 2x^2+10x=28 ?

The solutions to the equation 2x^2+10x=28 are x=2,x=-7

Find the zeros of 2x^2+10x=28

The zeros of 2x^2+10x=28 are x=2,x=-7

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2 x^{2} + 10 x = 28

x = 2, x = - 7

Solution steps

2 x^{2} + 10 x = 28

Move

28

to the left side

2 x^{2} + 10 x - 28 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 2 ( - 28 )}{2 \cdot 2}

1 0^{2} - 4 \cdot 2 (- 28) = 18

x_{1, 2} = \frac{- 10 \pm 18}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 10 + 18}{2 \cdot 2}, x_{2} = \frac{- 10 - 18}{2 \cdot 2}

x = \frac{- 10 + 18}{2 \cdot 2} : 2

x = \frac{- 10 - 18}{2 \cdot 2} : - 7

The solutions to the quadratic equation are:

x = 2, x = - 7

(26 + 8) (26 - 8)

(2^{8})^{2}

∣ 3 n + 8 ∣ - 4 = - 10

2 - 6 (5 x - 1) = - 202

2 x - 4 = 4

What are the solutions to the equation 2x^2+10x=28 ?
The solutions to the equation 2x^2+10x=28 are x=2,x=-7
Find the zeros of 2x^2+10x=28
The zeros of 2x^2+10x=28 are x=2,x=-7