What are the solutions to the equation 2x^2+20+10x=4x^2+28 ?

The solutions to the equation 2x^2+20+10x=4x^2+28 are x=1,x=4

Find the zeros of 2x^2+20+10x=4x^2+28

The zeros of 2x^2+20+10x=4x^2+28 are x=1,x=4

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2 x^{2} + 20 + 10 x = 4 x^{2} + 28

x = 1, x = 4

Solution steps

2 x^{2} + 20 + 10 x = 4 x^{2} + 28

Move

28

to the left side

2 x^{2} + 10 x - 8 = 4 x^{2}

Move

4 x^{2}

to the left side

- 2 x^{2} + 10 x - 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 ( - 2 ) ( - 8 )}{2 ( - 2 )}

1 0^{2} - 4 (- 2) (- 8) = 6

x_{1, 2} = \frac{- 10 \pm 6}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- 10 + 6}{2 ( - 2 )}, x_{2} = \frac{- 10 - 6}{2 ( - 2 )}

x = \frac{- 10 + 6}{2 ( - 2 )} : 1

x = \frac{- 10 - 6}{2 ( - 2 )} : 4

The solutions to the quadratic equation are:

x = 1, x = 4

X^{2} - 2 (X) (4) - (- 4)^{2} = 2 (X + 8)

(5 p + 3)^{2} = 8

2 x^{2} - 4 x + 17

2 x^{2} - 6 x = 4 x + 7

0.1 n^{2} + 100 n + 8 = 3600000

What are the solutions to the equation 2x^2+20+10x=4x^2+28 ?
The solutions to the equation 2x^2+20+10x=4x^2+28 are x=1,x=4
Find the zeros of 2x^2+20+10x=4x^2+28
The zeros of 2x^2+20+10x=4x^2+28 are x=1,x=4