What are the solutions to the equation 16x^2+10x-27=-6x+5 ?

The solutions to the equation 16x^2+10x-27=-6x+5 are x=1,x=-2

Find the zeros of 16x^2+10x-27=-6x+5

The zeros of 16x^2+10x-27=-6x+5 are x=1,x=-2

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16 x^{2} + 10 x - 27 = - 6 x + 5

x = 1, x = - 2

Solution steps

16 x^{2} + 10 x - 27 = - 6 x + 5

Move

5

to the left side

16 x^{2} + 10 x - 32 = - 6 x

Move

6 x

to the left side

16 x^{2} + 16 x - 32 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 \cdot 16 ( - 32 )}{2 \cdot 16}

1 6^{2} - 4 \cdot 16 (- 32) = 48

x_{1, 2} = \frac{- 16 \pm 48}{2 \cdot 16}

Separate the solutions

x_{1} = \frac{- 16 + 48}{2 \cdot 16}, x_{2} = \frac{- 16 - 48}{2 \cdot 16}

x = \frac{- 16 + 48}{2 \cdot 16} : 1

x = \frac{- 16 - 48}{2 \cdot 16} : - 2

The solutions to the quadratic equation are:

x = 1, x = - 2

\frac{x}{6} = 8

- 4.68 - 35.3

\frac{x - 2}{x ^{2} - 3 x - 10} \leq 0

8^{2} + 2^{2} = x^{2}

\frac{\frac{z ^{2} + 20 z + 96}{2 z + 3}}{( z + 8 ) ^{2}}

What are the solutions to the equation 16x^2+10x-27=-6x+5 ?
The solutions to the equation 16x^2+10x-27=-6x+5 are x=1,x=-2
Find the zeros of 16x^2+10x-27=-6x+5
The zeros of 16x^2+10x-27=-6x+5 are x=1,x=-2